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https://drive.google.com/drive/folders/1LZDyWdsgVJXxl18v2YCxq_LuVcwTBxzF?usp=share_link PROBABILITY INTRODUCTION Random Experiment An experiment, whose all possible outcomes are known in advance but the outcome of any specific performance can not predicted before the completion of the experiment, is known as random experiment. An example of random experiment might be tossing of a coin. Sample-space A set of all possible outcomes associated with same random experiment is called sample-space and is usually denoted by ‘S’. Consider the experiment of tossing a die. If we are interested in the number that shows on the top face, then sample space would be S1 = {1, 2, 3, 4, 5, 6} Event An event is a subset of sample – space. In any sample space we may be interested in the occurrence of certain events rather than in the occurrence of a specific element in the sample space. Simple Event If an event is a set containing only one element of the sample-space, then it is called a simple event. Compound Event A compound event is one that can be represented as a union of sample points. For instance, the event of drawing a heart from a deck of cards is the subset A ={heart} of the sample space S = {heart, spade, club, diamond}. Therefore A is a simple event. None the event B of drawing a red card is a compound event since B = {heart U diamond} = {heart, diamond}. PROBABILITY If a random experiment can result in any one of N different equally likely outcomes, and if exactly n of these outcomes favours to A, then the probability of event A, P (A) = . i.e. Remarks: If the probability of certain event is one, it doesn’t mean that event is going to happen with certainty! Infact we would be just predicting that, the event is most likely to occur in comparison to other events. Predictions depend upon the past information and of course also on the way of analysing the information at hand!! Similarly if the probability of certain event is zero, it doesn’t mean that, the event can never occur! Example -1: Two fair coins are tossed. What is the probability that atleast one head occurs? Sol: The sample space ‘S’ for this experiment is, S = {HH, HT, TH, TT} As the coin is fair all these outcomes are equally likely. Let ‘w’ be the weight assigned to any sample point then w + w + w + w = 1 ⇒ w = ¼ If ‘A’ is the event representing the event of atleast one head occuring, Then A = {HT, TH, HH} ⇒ P(A) = Example -2: In a single cast with two fair dice, what is the chance of throwing (i) two 4′s (ii) a doublet (iii) five - six (iv) a sum of 7 Sol: (i) There are 6 × 6 equally likely cases ( as any face of any die may turn up) ⇒ 36 possible outcomes. For this event, only one outcome (4-4) is favourable ∴ probability = 1/36). (ii)A doublet can occur in six ways {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}. Therefore probability of doublet = 6/36 = 1/6. (iii) Two favourable outcomes {(5, 6), (6, 5)} Therefore probability = 2/36 = 1/18 (iv) A sum of 7 can occur in the following cases {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} which are 6 in number. Therefore probability = 6/36 = 1/6 Example -3: Seven accidents occur in a week. What is the probability that they happen on the same day. Sol: Total no. of cases=Total no. of ways in which 7 accidents can happen in a week (or be distributed)= 77 Favourable no. of cases out of these = number of those in which all 7 happen on one day (any day of the week) = 7 ∴ Required. probability = Example -4: From a bag containing 5 white, 7 red and 4 black balls a man draws 3 balls at random, find the probability of being all white. Sol: Total number of balls in the bag =5+7+4= 16 The total number of ways in which 3 balls can be drawn is 16C3 = = 560 Thus sample space S for this experiment has 560 outcomes i.e. n(S) =560 Let E be the event of all the three balls being white. Total number of white balls is 5. So the number of ways in which 3 white balls can be drawn = 5C3= =10 Thus E has 10 elements of S, ∴ n(E) = 10 ∴ Probability of E, P(E) = == Example -5: From a set of 17 cards 1, 2, 3, ……16, 17, one is drawn at random. Find the probability that number on the drawn card would be divisible by 3 or 7. Sol: Numbers which are divisible by three are 3, 6, 9, 12, 15. Similarly numbers which are divisible by 7 are 7, 14. ⇒ Probability of the number written on the card to be divisible by 3 or 7 is = . Mutually Exclusive Events A set of events is said to be mutually exclusive if the occurrence of one of them precludes the occurrence of any of the other events. For instance, when a pair of dice is tossed, the events a sum of 4 occurs’, ‘a sum of 10 occurs’ and ‘a sum of 12 occurs’ are mutually exclusive. Simply speaking, if two events are mutually exclusive they can not occur simultaneously. Using set theoretic notation, if A1, A2,.. An be the set of mutually events then Ai ∩ Aj =φ for i ≠j and 1≤ i, j≤ n. Independent Events Events are said to be independent if the occurrence or non-occurrence of one does not affect the occurrence or non-occurrence of other. For instance, when a pair of dice is tossed, the events ‘first die shows an even number’ and ‘second die shows an odd number’ are independent. As the outcome of second die does not affect the outcome of first die. It should also be noted that these two events are not mutually exclusive as they can occur together. Remarks ∙ Distinction between mutually exclusive and independent events should be clearly made. To be precise, concept of mutually exclusive events is set theoretic in nature while the concept of independent events is probabilistic in nature. ∙ If two events A and B are mutually exclusive, they would be strongly dependent as the occurrence of one precludes the occurrence of the other. Exhaustive Event A set of events is said to be exhaustive if the performance of random experiment always result in the occurrence of atleast one of them. For instance, consider a ordinary pack of cards then the events ‘drawn card is heart’, drawn card is diamond’, ‘drawn card is club’ and ‘drawn card is spade’ is set of exhaustive event. In other words all sample points put together (i.e. sample space itself) would give us an exhaustive event. If ‘E’ be an exhaustive event then P(E) = 1. SET THEORETIC PRINCIPLES If ‘A’ and ‘B’ be any two events of the sample space then ∙ A∪B would stand for occurrence of atleast one of them. ∙ A ∩ B stands for simultaneous occurrence of A and B. ∙ (or A′) stands for non-occurrence of A ∙ (or A′ ∩ B′) stands for non-occurrence of both A and B. ∙ A ⊆ B stands for ‘the occurrence of A implies the occurrence of B’. ∙ If A and B are any two events, then P(A∪B) = P(A) + P(B) − P(A∩B) ∙ If A and B are mutually exclusive, P(A∪B) = P(A) + P(B). As in this case no sample point would be present in A∩B. ∙ P(A′) = 1 − P(A) As, A∪A′ = S and A and A′ are mutually exclusive ⇒ P(A∪A′) = P(A) + P(A′) = P(S) = 1 ⇒ P(A′) = 1 − P(A). ∙ P(A∩B′) = P(A) − P(A∩B) As A∩B′ and A∩B are mutually exclusive events and (A∩B′) ∪ (A∩B) = A ⇒ P(A∩B′) + P(A∩B) = P(A) ⇒ P(A∩B′) = P(A) − P(A∩B) Similarly, P(A′∩B) = P(B) − P(A∩B) ∙ P(A′∪B′) = 1 − P(A∩B) As, (A′ ∪ B′) ∪ (A ∩ B) = S ⇒ P(A′ ∪ B′) + P(A ∩ B) = 1 ⇒ P(A′ ∪ B′) = 1 − P(A∩B) Similarly, P(A′ ∩ B′) = 1 − P(A∪B) ∙ P(exactly one of A, B occurs) = P(A∩B′) + P(A′∩B) = P(A) + P(B) − 2P(A∩B) = P(A∪B) − P(A∩B) = P(A′ ∪ B′) − P(A′ ∩ B′) If A, B, C are any three events of the sample space, then ∙ P(A∪B∪C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A∩C) − P(B∩C) + P(A∩B∩C) ∙ P (Exactly one of A, B, C occurs) = P(A) + P(B) + P(C) − 2P(A∩B) − 2A(A∩C) − 2P(B∩C) + 3P(A∩B∩C) ∙ P(Exactly two of A, B, C occur) = P(A∩B) + P(B∩C) + P(A∩C) − 3P(A∩B∩C) ∙ P (at least two of A, B, C occur) = P(A∩B) + P(B∩C) + P(A∩C) − 2P(A∩B∩C) ∙ If A1, A2  , An are ′n` events, then P(A1 ∪ A2  ∪An) = +− + ∙ P (A ∪ B) ≥ max (P(A), P(B), P(A) + P(B) − 1) as A ⊆ A∪B ⇒ P(A) ≤ P(A∪B) Similarly, B ⊆ A ∪ B ⇒ P(B) ≤ P(A∪B) ⇒ P(A∪B) ≥ Max (P(A), P(B)) Also P(A∪B) = P(A) + P(B) − P(A ∩ B) P(A) + P(B) − 1 ≤ P(A∪B) ≤ P(A) + P(B) (As 0 ≤ P(A ∩ B) ≤ 1) Max (P(A) + P(B) − 1, P(A), P(B)) ≤ P(A ∪ B) ≤ P(A) + P(B) ∙ If out of m + n equally likely, mutually exclusive and exhaustive cases, m cases are favourable to an A event and n are not favourable to the an event A, m : n is called odds in favour of A, n : m is called odds against the event A. Example -6: Let A, B, C be three events. If the probability of occurring one event out of A and B is 1 - a, out of B and C is 1 - 2a, out of C and A is 1 - a and that of occurring three events simultaneously is a2, then prove that the probability that at least one out of A, B, C will occur is greater than or equal to 0.5. Sol: Probability that exactly one event out of A and B occur is p(A) + p(B) - 2p(A∩B) and probability that exactly one event out of B and C occur is p(B) + p(C) - 2p(B∩C) and so on. Now, p(A∪B∪C) = p(A) + p(B) + p(C) - p(A∩B) - p(A∩C) - p(B∩C) + p(A∩B∩C) =[p(A)+ p(B) - 2p(A∩B) + p(B) + p(C)-2p(B∩C)+p(C)+p(A)-2p(A∩C)] +p(A∩B∩C) = = a2 - 2a + Let a2 - 2a + = y ⇒ a2 - 2a + -y = 0 Since a is real, so 4 - 4( - y) ≥ 0 ⇒ y ≥ 1/2 . Example -7: From a pack of 52 cards two cards are drawn at random. Find the probability of the following events : (a) Both cards are of spade. (b) One card is of spade and one card is of diamond. Sol: The total number of ways in which 2 cards can be drawn = 52C2 == 26×51 = 1326. ∴ Number of elements in the space S are n(S) = 1326 (a) Let the event, that both cards are of spade, be denoted by E1, then n(E1) = Number of elements in E1 = Number of ways in which 2 cards can be selected out of 13 cards of spade = 13C2==78. ∴ Probability of E1 = P(E1)= . (b) Let E2 be the event that one card is of spade and one is of diamond then n(E2) = number of elements in E2 =number of ways in which one card of spade can be selected out of 13 spade cards and one card of diamond can be selected out of 13 diamond cards. = 13C1×13C1 = 13×13 =169 ∴ P(E2) = . Example -8: A has 3 shares in a lottery containing 3 prizes and 6 blanks; B has one share in a lottery containing one prize and 2 blanks. Compare their chances of success. Sol: Total number of tickets in the first lottery = 3 + 6 =9 A may select any three tickets out of these 9 tickets. Therefore, the number of elements in the sample space S is given by =n(S) =9C3 = =84 Let E1 be the event of winning the prize in the first lottery by A .So, is he event of not winning the prize of A, Then number of elements in is given by n() = number of ways of selecting 3 tickets out of six blank tickets = 6C3 = =20 ∴The probability of not winning the prize by A is () = Since P(E1) +P() =1 ∴ P(E1) = 1-P() Or P(E1)= 1- Now for B n(S) = number of ways of selecting 1 ticket out of 3 tickets. Or n(S) = 3C1 = 3. Let E2 be the event of winning the prize by B. ∴ n(E2) = number of ways of selecting one ticket out of 1 prize ticket =1C1 =1 ∴ The probability of winning the prize by B. P(E2) = ∴ The ratio of the probabilities of winning the prizes by A and B =. CONDITIONAL PROBABILITY The probability of occurrence of an event B when it is known that some event A has occurred is called a condition probability and is denoted by P(B/A). The symbol P(B/A) is usually read '‘the probability that B occurs given that A occurs” or ”simply probability of B, given A”. Consider two events ‘A’ and ‘B’ of sample-space S. When it is known that event ‘A’ has occurred, it means that sample space would reduce to the sample points representing event A. Now for P(B/A) we must look for the sample points representing the simultaneous occurrence of A and B i.e. sample points in A∩B. ⇒ P(B/A) = Thus P(B/A) = , where 0 < P(A) ≤ 1 Similarly, P(A/B) = 0 < P(B) ≤ 1 Hence, P(A ∩ B) = Consider the event ‘B’ of getting a ‘4’ when a fair die is tossed. Now suppose that it is known that toss of die resulted in a number greater than 3 (say event A). And we have to obtain P(B/A) i.e. the probability of getting a ‘4’ given that a number greater than 3 has occurred. Clearly A = {4, 5, 6}, B = {4} ⇒ P(B/A) = also P(A∩B) = and P(A) = ⇒ P(B/A) = = Example -9: The probability that a married man watches a certain T.V. show is 0.4 and the probability that a married woman watches the show is 0.5. The probability that a man watches the show, given that his wife does, is 0.7. Find (a) the probability that married couple watch the show (b) the probability that a wife watches the show given that her husband does. (c) the probability that atleast one person of a married couple will watch the show. Sol: Let ‘H’ be the event that married man watches the show and ‘W’ be the probability that married woman watch the show, ⇒ P(H) = 0.4, P(W) = 0.5, P(H/W) = 0.7 (a) P(H∩W) = P(W)⋅P(H/W) = 0.5 × 0.7 = 0.35 (b) P(W/H) = (c) P(H∪W) = P(H) + P(W) − P(H∩W) = 0.4 + 0.5 − 0.35 = 0.55. Example -10: Consider the sample space ‘S’ representing the adults in a small town who have completed the requirements for a college degree. They have been categorised according to sex and employment as under: Employed Unemployed Male 460 40 Female 140 260 A employed person is selected at random. Find the probability that chosen one is male. Sol: Let M be the event that man is chosen and E be the event that chosen one is employed. From the concept of reduced sample space we immediately get P(M/E) = Also, P(E) = P(E∩M) = ⇒ P(M/E) = Example -11: A bag contains 3 white balls and 2 black balls, another contains 5 white and 3 black balls. If a bag is chosen at random and a ball is drawn from it. What is the probability that it is white. Sol: The probability that the first bag is chosen is 1/2 and the chance of drawing a white ball from it is 3/5. ∴ Chance of choosing the first bag and drawing a white ball is 1/2.3/5 Similarly, the chance of choosing the second bag and drawing a white ball is 1/2.5/8. ∴ The chance of randomly choosing a bag and drawing a white ball is = (Mutually exclusive cases) = 49/80. Example -12: Find the probability that an year chosen at random has 53 Sundays. Sol: Let P(L) → be the probability that an year chosen at random is leap year P(L) = 1/4, ∴ P = 3/4 Let P(S) → be the probability that an year chosen at random has 53 sundays. ∴ P(S) = P(L). P(S/L) + P().P(S/) Now P(S/L) prob. that a leap year has 53 Sundays. A leap year has 366 days, 52 weeks + the remaining 2 days may be Sunday-Monday, M-T,T-W, W-Th, Th-F, F-Sat or Sat –S, Out of the 7 possibilities 2 are favourable ∴ P(S/L) = . Similarly P(S/) = ∴ P(S) = Independent Events: Two events A and B are said to be independent if occurrence or non-occurrence of one does not affect the occurrence or non-occurrence of the other, i.e. P(B/A) = P(B), P(A) ≠ 0 similarly P(A/B) = P(A), P(B) ≠ 0 ⇒ P(B/A) = ⇒ P(A∩B) = P(A)⋅P(B) If the events are not independent, they are said to be dependent. Remarks: ∙ If P(A) = 0 ⇒ for event ‘B’, 0 ≤ P(A∩B) ≤ P(A). ⇒ P(A∩B) = 0, thus P(A∩B) = P(A)⋅P(B) = 0. Hence an impossible event would be independent of any other event. ∙ Distinction between independent and mutually exclusive events must be carefully made. Independence is a property of probability whereas the mutual exclusion is a set-theoretic concept. If A and B are two mutually exclusive and possible events of sample space ‘S’ then P(A) > 0, P(B) > 0 and P(A∩B) = 0 ≠ P(A)⋅P(B) so that A and B can’t be independent. Infact P(A/B) = 0 similarly P(B/A) = 0. Consequently, mutually exclusive events are strongly dependent. ∙ Two events A and B are independent if and only if A and B′ are independent or A′ and B are independent or A′ and B′ are independent. We have P(A∩B) = P(A)⋅P(B) Now P(A∩B′) = P(A) − P(A∩B) = P(A) − P(A)⋅P(B) = P(A) (1−P(B)) = P(A)⋅P(B′) Thus A and B′ are independent. Similarly P(A′∩B) = P(B) − P(A∩B) = P(B)⋅P(A′) Finally P(A′ ∩ B′) = 1 − P(A∪B) = 1 − P(A) − P(B) + P(A∩B) = 1 − P(A) − P(B) + P(A)⋅P(B) = (1−P(A)) − P(B) (1−P(A)) = (1−P(A)) (1−P(B)) = P(A′)⋅P(B′) Thus A′ and B′ are also independent. Example -13: An event A1 can happen with probability p1 and event A2 can happen with probability p2. What is the probability that (i) exactly one of them happens (ii) at least one of them happens(Given A1 and A2 are independent events). Sol: (i) The probability that A1 happens is p1 ∴ The probability that A1 fails is 1 - p1 Also the probability that A2 happens is p2 Now, the chance that A1 happens and A2 fails is p1(1 - p2) and the chance that A1 fails and A2 happens is p2(1-p1) ∴ The probability that one and only one of them happens is p1(1 - p2) + p2(1 - p1) = p1 + p2 - 2p1p2 (ii) The probability that both of them fail = (1 - p1) (1 - p2). ∴ probability that atleast one happens=1 -(1 - p1) (1 - p2)=p1+p2 - p1 p2 Mutual Independence and Pairwise Independence: Three events A, B, C are said to be mutually independent if, P(A∩B) = P(A)⋅P(B), P(A ∩ C) = P(A)⋅P(C), P(B ∩ C) = P(B)⋅P(C) and P(A∩B∩C) = P(A)⋅P(B)⋅P(C) These events would be said to be pairwise independent if, P(A∩B) = P(A)⋅P(B), P(B∩C) = P(B)⋅P(C) and P(A∩C) = P(A)⋅P(C). Thus mutually independent events are pairwise independent but the converse may not be true. Example -14: A lot contains 50 defective and 50 non-defective bulbs. Two bulbs are drawn at random, one at a time with replacement. The events A, B, C are defined as under; A = {the first bulb is defective} B = {the second bulb is non defective} C = {the two bulbs are both defective or both are non-defective}. Categorise the events A, B, C to be pairwise independent or mutually independent. Sol: Let Di (Ni) denote the occurrence of a defective (non-defective) bulb at the ith draw (i = 1, 2). Now sample-space ‘S’ is; S = {D1 D2, D1 N2, N1 D2, N1 N2} ⇒ P(Di) = (i = 1, 2) A = {D1 D2, D1 N2}, B = {D1 N2, N1 N2}, C = {D1 D2, N1 N2} ⇒ A∩B = {D1 N2} ⇒ A∩C = {D1 D2} ⇒ B∩C = {N1 N2} and A ∩ B ∩ C = φ P(A) = P(D1 D2) + P(D1N2) = P(B) = P(D1 N2) + P(N1 N2) = P(C) = P(D1 D2) + P(N1N2) = P(A∩B) = P(D1 N2) = = P(A)⋅P(B) P(B∩C) = P(N1 N2) == P(B)⋅P(C) P(A∩C) = P(D1 D2) = = P(A)⋅P(C) And P(A ∩ B ∩ C) = 0 ≠ P(A) ⋅ P(B) ⋅ P(C) ⇒ Events are not mutually independent but are pairwise independent. TOTAL PROBABILITY THEOREM AND BAYES’ THEOREM Partition of Sample-Space: Consider a sample space ‘S’. Let A1, A2,  An be the set of mutually exclusive and exhaustive set of sample-space S. These A1, A2, , An events are said to partition the Sample-space in to ‘n’-parts. We have Ai ∩ Aj ≠ φ for i ≠ j, i ≤ i, j ≤ n And Total Probability Theorem: Let ‘A’ be any event of S. We can write A = (A1 ∩ A) ∪ (A2 ∩ A)  ∪(An ∩ A). As A1, A2  An are mutually exclusive, (A1 ∩ A), (A2 ∩ A), , (An ∩ A) would also be mutually exclusive. ⇒ P(A) = P(A1 ∩ A) + P(A2 ∩ A) +  + P(An ∩ A) = P(A1)⋅P(A/A1) + P(A2)⋅P(A/A2) +  + P(An)⋅P(A/An) ⇒ P(A) = This is known as the total probability of the event A. Remark: ∙ P(A/Ai) gives up the contribution of Ai in the occurrence of A. Example -15: One bag contains four white balls and three black balls and a second bag contains three white balls and five black balls. One ball is drawn from the first bag and placed unseen in the second bag. What is the probability that a ball now drawn from second bag is black ? Sol: Bag – I Bag – II 4 w 3 w 3 B 5 B Let A1 be the event that a white ball is transferred from bag – I to bag – II and A2 be the event that a black ball is transferred from bag – I to bag –II. P (A1) =, P(A2) = Let 'A' be the probability the finally a black ball is drawn from the second bag P(A/A1) = , P(A/A2) = . Now from total probability theorem we get, P(A) = P(A1) . P(A/A1) + P(A2) . P(A/A2) = Example -16: A real estate man eight master keys to open several new homes. Only one master key will open any given house. If 40% of these homes are usually left unlocked, what is the probability that the real estate man can get into a specific home if he selects three master keys at random before leaving the office ? Sol: Let A1 and A2 be the events that the specific home is left unlocked, and is left locked respectively. ⇒ P(A1) = 0.4 , P (A2) = 0.6 Let 'A' be the event that real estate man get into the specific home. P(A/A1) = 1, P(A/A2) = . ⇒ P(A) = P(A1) P(A/A2) + P(A2) P(A/A2) = (0.4) (1) + (0.6) (3/8) = = . Bayes’ Theorem: This theorem at times is also called inverse probability theorem. Let us consider any event ‘A’ of sample space ‘S’ (as in the previous section). This event would have occurred due to the different causes (or due to the occurrence of any of the event A1, A2,  An). Now, let us say that event A is found to have occurred and we have to find the probability that it has occurred to the occurrence of cause, say Ai. That means we are interested in P(Ai/A). These types of problems are solved with the help of Bayes’ theorem. From total probability theorem we get P(A) = Also, P(Ai/A) = ⇒ P(Ai/A) = This result is known as Bayes’ theorem. Example -17: A bag 'A' contains 2 w and 3 red balls, a bag 'B' contains 4 w and 5 black balls. A bag is selected randomly and a ball is drawn from it. Drawn ball is observed to be white. Find the probability that bag 'B' was selected. Sol: Bag A Bag B 2 w, 3R 4 w, 5B Let A1 be the event that bag 'A' is selected and A2 be the event that bag B is selected. P(A1) = P(A2) =1/2 Let 'A' be the event that a white ball is drawn from the selected bag. ⇒ P(A/A1) = 2/5, P(A/A2) = P(A) = P(A1) . P(A/A1) + P(A2) . P(A/A2) = Finally, P(A2/A) = = Example -18: A card from a pack of 52 cards is lost. From the remaining cards, two cards are drawn and are found to be spades. Find the probability that missing card is also a spade. Sol: Let A1 be the event that missing card is spade and A2 be event that missing card is non-spade. ⇒ P(A1) = , P(A2) = Let 'A' be event that 2 spade cards are drawn from the remaining cards, ⇒ P(A) = P(A1) ⋅ = Now Example -19: A company manufactures T. Vs at two different plants A and B. Plant 'A' produces 80 % and B produces 20 % of the total production. 85 out of 100 TVs produced at plant A meet the quality standards while 65 out of 100 T.Vs produced at plant B meet the quality standard. A T.V. produced by the company is selected at random and is not found to meeting the quality standard. Find the probability that selected T.V. was manufactured by the plant B. Sol: Let B1 (B2) be the event that plant A(B) produce a T.V. that does not meet the quality standard. P(B1) = 1 - P(B2) = 1 - Let A1(A2) be the event that selected T.V. is produced by plant A(B) ⇒ P(A1) = 0.8, P(A2) = 0.2 Let 'A' be the event that selected T.V. does meet the quality standards. ⇒ P(A/A1) = P(B1) = and P(A/A2) = P(B2) = P(A) = P(A1). P(A/A1) + P(A2). P(A/A2) = 0.8 × + (0.2) = ⇒ P(A/A2) = = = . BINOMIAL TRIALS AND BINOMIAL DISTRIBUTION Consider a random experiment whose outcomes can be classified as success or failure. It means that experiment results in only two outcomes E1(success) or E2 (failure). Further assume that experiment can be repeated several times, probability of success or failure in any trial are p and q (p + q = 1) and don’t vary from trial to trial and finally different trials are independent. Such a experiment is called Binomial experiment and trials are said to be binomial trials. For instance tossing of a fair coin several times, each time outcome would be either a success (say occurrence of head) or failure (say occurrence of tail). A probability distribution representing the binomial trials is said to binomial distribution. Let us consider a Binomial experiment which has been repeated ‘n’ times. Let the probability of success and failure in any trial be p and q respectively and we are interested in the probability of occurrence of exactly ‘r’ successes in these n trials. Now number of ways of choosing ‘r’ success in ‘n’ trials = nCr. Probability of ‘r’ successes and (n−r) failures is pr⋅qn−r. Thus probability of having exactly r successes = nCr⋅pr⋅qn−r Let ‘X’ be random variable representing the number of successes, then P(X = r) = nCr⋅pr⋅qn−r (r = 0, 1, 2,  , n) 1 = (p + q)n = nC0p0qn + nC1p1qn –1 + nC2p2qn –2 + ……. + nCrprqn –r + ……. + nCnpn X → Number of success 0, 1, 2, …….., r, ……….. n Remarks: ∙ Probability of utmost ‘r’ successes in n trials = ∙ Probability of atleast ‘r’ successes in n trials = ∙ Probability of having 1st success at the rth trial = p⋅qr−1 Example -20: A die is thrown 7 times. What is the chance that an odd number turns up (i) exactly 4 times (ii) at least 4 times. Sol: Probability of success = ⇒ p = (i) For exactly 4 successes required probability = 7C4. (ii) For atleast 4 successes required probability = 7C4 = SOLVED PROBLEMS SUBJECTIVE Problem -1: A manufacturing company uses an acceptance scheme on production items before they are shipped. The plan is a two-stage one. Boxes of 25 are readied for shipment and a sample of 3 are tested for defectives. If any defectives are found, the entire box is sent for 100% screening. If no defectives are found, the box is shipped. (a) what is the probability that a box containing 3 defectives will be shipped? (b) what is the probability that a box containing only one defective will be sent back for screening? Sol: (a) A box containing 3 defectives will be shipped, if no defective item is included in the sample. ⇒ Required probability = (b) In this case sample taken must contain the defective item, ⇒ Required probability = Problem -2: In a multiple-choice question there are four alternative answers, of which one or more than one are correct. A candidate will get marks on the question only if he ticks all the correct answers. The candidate decides to tick answers randomly. If he is allowed up to three chances to answer the questions, find the probability that he will get marks on it. Sol: Total number of ways of answering the question, = 4C1 + 4C2 + 4C3 + 4C4 = 15 Now out of these, only one combination is correct. ⇒ Probability of answering the question correctly at the first trial = Now, Probability of answering the question correctly at the second trial, = (with the assumption that student doesn’t repeat the mistake of first trial) Similarly, probability of answering the question correctly at the third trial, = ⇒ Probability of required event = Problem -3: If p and q are chosen randomly from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} with replacement, find the probability that the roots of the equation x2+ px + q = 0 are real and distinct. Sol: For roots of x2 + px + q = 0 to be real and distinct we must have, p2 > 4q ⇒ p ≥ 3 as q ≥ 1 There are following possibilities; q 4q p Total cases 1 4 3,4,5,6,7,8,9,10 8 2 8 3,4,5,6,7,8,9,10 8 3 12 4,5,6,7,8,9,10 7 4 16 5,6,7,8,9,10 6 5 20 5,6,7,8,9,10 6 6 24 6,7,8,9,10 6 7 28 6,7,8,9,10 5 8 32 6,7,8,9,10 5 9 36 7,8,9,10 4 10 40 7,8,9,10 4 Total number of favourable cases 59 Total number of ways in which p and q can be chosen with replacement = 10. 10 = 100 ⇒ Required probability = . Problem -4: The odds in favour of standing first of three students appearing at an examination are 1 : 2, 2 : 5 and 1 : 7 respectively. What is the probability that either of them will stand first. Sol: Let the three students be P, Q and R Let A, B, C denote the events of standing first of the three students P, Q, R respectively Given, odds in favour of A = 1 : 2 Odds in favour of B = 2 : 5 And odds favour of C = 1 : 7 ∴ P (A) = , P (B) = , P (C) = since events A, B and C are mutually exclusive ∴ P (A ∪ B ∪ C) = P (A) + P (B) + P (C) = Problem -5: A bag contains 5 white and 8 red balls. Two drawns of 3 balls each are made without replacement. What is the probability that the first draw gives 3 white balls and second draw gives 3 red balls? Sol: Let A = the event of drawing 3 white balls in the first draw And B = the event of drawing 3 red balls in the second draw Then A ∩ B = the event of drawing 3 white balls in the draw and 3 red ball in the second draw Now P (A) = When 3 white balls are drawn in the first draw, number of balls left = 10 And number of red balls left = 8 ∴ now required probability, P (A ∩ B) = P (A) Problem -6: Three students appear at an examination of mathematics. The probability of their success are respectively. Find the probability of success of at least two. Sol: Let E1 = the event of success of the first student. E2 = the event of success of the second student. E3 = the event of success of the third student. Let A = E1 ∩ E2 ∩ E’ 3 = the event that the first and second student succeed and the third fails B = ∩ E2 ∩ E3 , C = E1 ∩ ∩ E3 D = E1 ∩ E2 ∩ E3 According to question, P (E1) = , P (E2) = , P (E3) = ∴ P = , P = , P = Clearly, A ∪ B ∪ C ∪ D = the event of success of at least two students Since, A, B, C, D are mutually exclusive events and E1, E2, E3 are independent events. ∴ required probability, P (A ∪ B ∪ C ∪ D) = P (A) + P (B) + P (C) + P (D) = P ( E1 ∩ E2 ∩ ) + P (∩ E2 ∩ E3) + P ( E1 ∩ ∩ E3) + P (E1 ∩ E2 ∩ E3) = P ( E1) . P ( E2) . P () + P () . P ( E2) . P ( E3) + P ( E1) . P () . P (E3) + P (E1) . P(E2) . P (E3) = = . Problem -7: There are two purses. The first contains 9 fifty-paise coins and a one-rupee coin, while the second purse has 10 fifty-paise coins. Nine coins are transferred from the first purse to second randomly. Then nine coins are transferred from the second purse to the first randomly. Find the probability of finding a one-rupee coin in the first purse after these transfers. Sol: Let E1 = the event that one rupee coin is transferred from first purse to second purse when 9 coins are transferred. Let E2 = the event that one rupee coin is not transferred from first purse to second purse when 9 coins are transferred. E = the event that one rupee coin is found in the first purse the second transfer. Let A1 = E1 ∩ E, A2 = E2 ∩ E Let A = A1 ∪ A2 Since E1 and E2 are mutually exclusive events, therefore A1 and A2 are also mutually exclusive. Therefore, required probability, P (A) = P (A1) + P (A2) = P (E1 ∩ E) + P (E2 ∩ E) Hence P (A) = P (E1) . P (E/E1) + P (E2) . P (E/E2) ..(1) Now, P (E/E2) = probability that one rupee coin will be transferred from second purse to the first purse when one rupee coin and 8 fifty paise coins are transferred from first purse to the second purse = P (E/E2) = probability that one rupee coin will be in the first purse when it has been transferred to the second purse = 1 From (1), P (A) = = Problem -8: A tosses 2 fair coins and B tosses 3 fair coins. The game is won by the person who throws greater number of heads. In case of a tie, the game is continued under identical rules until someone wins the game. Find the probability of A winning the game. Sol: For a specific game let Ai (Bi) denote the number of heads obtained by A(B) is i when he tosses two (three) fair coins. A will win a particular game in following ways, A1 and B0 occur; A2 and B0 occur, or A2 and B1 occur. If AW be the event that A wins the specific game, then P(AW) = P(A1 ∩ B0) + P(A2 ∩ B0) + P(A2 ∩ B1) = P(A1)⋅P(B0) + P(A2)⋅P(B0) + P(A2)⋅P(B1) = Now game results in a tie if A0 and B0 occur or A1 and B1 occur or A2 and B2 occur. If T be event representing a tie; ⇒ P(T) = P(A0)⋅P(B0) + P(A1) ⋅P(B1) + P(A2)⋅P(B2) = Now required probability, = P(AW)+P(T)⋅P(AW)+(P(T))2⋅P(AW)+  = Problem -9: An unbiased coin is tossed. If the result is a head, a pair of unbiased dice is rolled and the number obtained by adding the numbers on the two faces is noted. If the result is a tail, a card from a well-shuffled pack of eleven cards numbered 2, 3, 4,, 12 is picked and the number on the card is noted. Find the probability that noted number is either 7 or 8. Sol: Let A1(A2) be the event that tossing of coin results in head (tail). And A be the event that noted number is 7 or 8. P(A1) = P(A2) = P(A/A2) = P(A/A1) = (As 7 can result in, (1, 6), (2,5), (3,4), (6,1), (5,2), (4, 3) i.e. 6 ways and 8 can result in (2, 6), (3,5), (4,4), (6,2), (5,3) i.e. 5 ways) Now P(A) = P(A1)⋅P(A/A1) + P(A2)⋅P(A/A2) = = Problem -10: In a single throw with 3 dice, what is the chance of throwing (i) one-two-three (ii) eleven (iii) less than eleven (iv) more than ten. Sol: The total no. of possible cases is 63 = 216. (i) Out of these 216, 1 - 2 - 3 occurs in 3! ways. Probability = 3!/63 = 1/36. (ii) The no. of ways in which 11 is thrown is equal to the coeff. of x11 in (x + x2 +....+ x6)3 = 27. Probability = 27/216 = 1/8 (iii) The favourable no. of cases = sum of the coefficients of x3, x4, ... x10 in the expansion of (x + x2 + .... x6)3 = the coeff. of x10 in (x + x2+....+x6)3 (1 + x + x2+...) = the coeff. of x10 in x3(1 - x6)3 (1 - x)-4 = the coeff. of x in (1 - 3x6) (1 + 4x + x2 + x3 + .....) = = 120 - 12 = 108 Probability of throwing less than 11 = (Note : Here we made use of the fact that if S = a0 + a1x + a2x2 + .... then S(1 + x + x2 + ....) = a0 + (a0 + a1)x + (a0 + a1 + a2)x2 + .......) (iv) The total no. of cases are those which are either less than 11 or greater than equal to 11 (i.e. greater than 10) Thus required probability = 1 - prob. of less than 11 = 1 - . OBJECTIVE 1. A number of six digits is written down at random. Probability that sum of digits of the number is even is (A) 1/2 (B) 3/8 (C) 3/7 (D) none of these Sol: As we are considering all possible six digit numbers, half of these would have sum of their digits to be even and half of these would have sum of their digits to be odd. Hence the required probability would be 1/2. 2. Two small squares on a chess board are chosen at random. Probability that they have a common side is, (A) 1/3 (B) 1/9 (C) 1/18 (D) none of these Sol: There are 64 small squares on a chessboard. ⇒ Total number of ways to choose two squares = 64C2 = 32.63 For favourable ways we must chosen two consecutive small squares for any row or any columns ⇒ Number of favorable ways = (7.8)2 ⇒ Required probability = 3. A bag contains 17 tickets numbered 1 to 17. A ticket is drawn and replaced, then one more ticket is drawn and replaced. Probability that first number drawn is even and second is odd is (A) (B) (C) (D) none of these Sol: Let A be the event that first drawn number is even and B be event that the second drawn number is odd. Total number of even numbers = = 8 ⇒ P(A) = and P(B) = ⇒ P(A ∩B) = 4. A and B play a game of tennis. The situation of the game is as follows; if one scores two consecutive points after a deuce he wins; if loss of a point is followed by win of a point, it is deuce. The chance of a server to win a point is 2/3. The game is at deuce and A is serving. Probability that A will win the match is, (serves are changed after each game) (A) 3/5 (B) 2/5 (C) 1/2 (D) 4/5 Sol: Let us assume that 'A' wins after n deuces, n ∈ [( 0, ∞ ) Probability of a deuce = (A wins his serve then B wins his serve or A loses his serve then B also loses his serve) Now probability of 'A' winning the game = 5. A die is thrown three times and the sum of three numbers obtained is 15. The probability of first throw being 4 is (A) 1/18 (B) 1/5 (C) 4/5 (D) 17/18 Sol: If first throw is four, then sum of numbers appearing on last two throws must be equal to eleven. That means last two throws are (6, 5) or (5, 6) Now there are 10 ways to get the sum as 15. [(5, 5, 5) (4, 5, 6) (3,6,6)] ⇒ Required probability = 6. Fifteen coupons are numbered 1, 2, 3, …. 15. Seven coupons are selected at random one at a time with replacement. The probability that the largest number appearing on the selected coupon is 9, is (A) (B) (C) (D) none of these Sol: Total ways = 157 For favorable ways, we must 7 coupons numbered from 1 to 9 so that '9' is selected atleast once. Thus total number of favorable ways are, 97 - 87 ⇒ Required probability = 7. A fair die is thrown until a score of less than 5 points is obtained. The probability of obtaining not less than 2 points on the last thrown is (A) 3/4 (B) 5/6 (C) 4/5 (D) 1/3 Sol: Score less than 5 means the occurrence of 1, 2, 3, or 4. Now on the last throw we should not obtain a score less then 2 i.e. one. Clearly the favourable outcomes are 2, 3 or 4. Thus the required probability = 3/4 8. A fair coin is tossed a fixed number of times . If the probability of getting 7 heads is equal to getting 9 heads, then the probability of getting 2 heads is, (A) 15/28 (B) 2/15 (C) 15/213 (D) none of these Sol: Let coin was tossed 'n' times and X be the random variable representing the number of head appearing in 'n' trials. P(X = 7) = P (X = 9) ⇒ n C7(1/2)7 (1/2)n-7 = nC9(1/2)n-9 × (1/2)9 ⇒ nC7 = nC9 ⇒ n = 16 Now P (X = 2) = 16C2(1/2)2 (1/2)14 = = . 9. A fair die is tossed eight times. Probability that on the eighth throw a third six is observed is, (A) 8C3 (B) (C) (D) none of these Sol: Third six occurs on 8th trial. It means that in first 7 trials we must exactly 2 sixes and 8th trial must result in a six. ⇒ Required probability = 7C2 . (1/6)2. (5/6)5. (1/6) =. 10. If the papers of 4 students can be checked by any one of the 7 teachers, then the probability that all the 4papers are checked by exactly 2 teachers is; (A) 2/ 7 (B) 32/ 343 (C) 12/ 49 (D) None of these Sol: Total number of ways in which 4 papers can be distributed among 7 teachers =74 Now exactly 2 teachers out of 7 can be chosen in 7C2 ways. And total number of ways in which 4 papers can be given to these 2 teachers ( each one getting atleast one) = (24 –2 ) =14 ⇒ Total number of ways in which exactly 2 teachers check all four papers = 7C2. 14= 21 .14 ⇒ Required probability = . 11. A and B each throw a dice. The probability that A's throw is not greater than B's is (A) 1/6 (B) 5/12 (C) 1/2 (D) 7/12 Sol: A ≤ B P (A > B) B draw A draw Total numbers 1 2, 3, ……,6 5 2 3, 4, …., 6 4 3 4, 5, 6 3 4 5, 6 2 5 6 1 Total = 15 P (A > B) = P (A ≤ B) = 1 – = . 12. A speaks the truth in 70% cases and B speaks the truth in 80% cases. The probability that they will contradict each other in describing a single event is (A) 0.38 (B) 0.56 (C) 0.62 (D) 0.94 Sol: P (A) = , P (B) = P (C) = = = 0.38 13. If A and B are two events, the probability that exactly one of them occurs is given by (A) P(A) + P(B) – 2 P(A∩B) (B) P(A∩B′) – P(A′∩B) (C) P(A∪B) + P(A∩B) (D) P(A′) + P(B′) + 2 P(A′∩B′) Sol: P(A ∪ B) = P(A) – P(AB) + P(B) – P(AB) = P(A) + P(B) – 2P(AB) 14. If a number of two digit is formed with the digit 2, 3, 5, 7, 9 without repetition of digits, then the probability that the number formed 35 is (A) (B) (C) (D) none of these Sol: Required probability = . 15. If A and B are independent events such that P (A) = 0.3, P (A ∪ ) = 0.8, then P (B) equals to (A) (B) (C) (D) Sol: P (A) + P () – P (A) P () = 0.8 0.3 + (1 – x) – 0.3(1 – x) = 0.8 (1 – x)(0.7) = 0.5 1 – x = ⇒ x = 1 – = 16. For three events A, B and C, P (exactly one of the events A or B occurs) = P (exactly one of the events B or C occurs) = P (exactly one of the events C or A occurs) = p and P (all the three events occur simultaneously) = p2, where 0 < p< 1/ 2. Then the probability of at least one of the three events A, B and C occurring is (A) (B) (C) (D) Sol: Given P (A) + P (B) – 2P (AB) = P P (B) + P (C) – 2P (BC) = P P (C) + P (A) – 2P (AC) = P 2 [P (A) + P (B) + P (C) – P (AB) + P (BC) + P (CA)] = and P (ABC) = P2 Hence P (A ∪ B ∪ C) = + P2 = . 17. A four figure number is formed of the figures 1,2,3,5, with no repetitions. The probability that the number is divisible by 5 is (A) 3/4 (B) 1/4 (C) 1/8 (D) None of these Sol: Required probability = 18. Three natural numbers are taken at random from the set A= {x : 1 ≤ x ≤ 100, x∈N. then the probability of that AM of the number taken is 25 (A) (B) (C) (D) none of these Solution : x1 + x2 + x3 = 75 coefficient x75 in (x1 + x2 + …… + x100)3 = 74C2 Hence probability = . Exercise # 1 19. A bag contains 5 white and 3 black balls. Two balls are drawn at random. The probability that the drawn balls are of different colours, is equal to (Ref. P.K.SHARMA Pg.A8.1 Q.no.1) (A) (B) (C) (D) Sol: (B) Drawn balls should be such that one of them is white and another is black. Total ways of drawing two balls = 8C2 = 28. Total number of favourable ways = 5C1 ⋅ 3C1 = 15 Thus, required probability = 20. The probability that at least one of the events A and B happens is . Probability of their simultaneous happening is . Value of P + P is equal to (Ref. P.K.SHARMA Pg.A8.1 Q.no.2) (A) (B) (C) (D) Sol: (C) P(A ∪ B) = , P(A ∩ B) = Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 2 – P(A ∩ B) – P() – P() ⇒ P() + P() = 2 – P(A ∩ B) – P(A ∪ B) = 2 – 21. One number is selected at random from first two hundred positive integers. The probability that it is divisible by 6 or 8, is equal to (Ref. P.K.SHARMA Pg.A8.1 Q.no.3) (A) (B) (C) (D) Sol: (A) Total number of integers that are divisible by six = = 38 Total number of integers that are divisible by eight = = 28 Total number of integers that are divisible by six as well as by eight i.e. by 24 = = 8 Thus total number of integers that are divisible by 6 or 8 = 38 + 28 – 8 = 58 Thus, required probability = 22. Two fair and ordinary dice are rolled simultaneously. The probability of getting the sum of outcomes of the dice as a multiple of 4, is equal to (Ref. P.K.SHARMA Pg.A8.1 Q.no.4) (A) (B) (C) (D) Sol: (B) In the case sum can be 4 or 8 or 12. Total ways of getting the sum as 4 = 3 (namely (2, 2), (1, 3), (3, 1)). Total ways of getting the sum as 8 = 5 (namely (2, 6), (6, 2), (3, 5), (5, 3), (4, 4)) Total ways of getting the sum as 12 = 1 (namely (6, 6)) Thus total ways of getting the sum as multiple of four = 3 + 5 + 1 = 9 Thus, required probability = 23. The probabilities of happening of the events A and B are and respectively. If the probability of happening of A and B simultaneously is , then probability of neither A nor B happening, is equal to (Ref. P.K.SHARMA Pg.A8.2 Q.no.7) (A) (B) (C) (D) None of these Sol: (A) P(A) = , P(B) = , P(A ∩ B) = . Now, P( ∩ ) = P() + P() – P( ∪ ) = 1 – P(A) + 1 – P(B) – (1 – P(A ∩ B)) = 1 – – 24. A five digit number (having all different digits) is formed using the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9. The probability that the formed number either begins or ends with an odd digit, is equal to (Ref. P.K.SHARMA Pg.A8.2 Q.no.8) (A) (B) (C) (D) Sol: (A) Total formed numbers that begin with a odd digit = 5C1 ⋅ 8P4 = 5(8) (7) (6) (5) Total formed numbers that end with a odd digit = 5C1 ⋅ 8P4 = (8) (7) (6) (5) Total formed number that begin with an odd digit and also end with an odd digit = 5C2 ⋅ 2! ⋅ 7P3 = 5 ⋅ (4) (7) (6) (5) Thus total formed numbers that begin with an odd digit or end with an odd digits is equal to 5 ⋅ 7 ⋅ 6 ⋅ 60 Total formed numbers = 9P5 = 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 Thus, required probability = 25. The probabilities of solving a problem by students A, B and C independently are , and respectively. If they start solving the given problem. independently, then the probability that atleast two of them will solve the problem successfully, is equal to (Ref. P.K.SHARMA Pg.A8.2 Q.no.9) (A) (B) (C) (D) Sol: (C) Probability that A and B can solve but ‘C’ is unable to solve = Probability that A and C can solve but ‘B’ is unable to solve = Probability that B and C can solve but ‘A’ is unable to solve = Probability that all of them can solve the problem = Thus required probability that atleast two of them can solve the problem = 26. The probability that the persons and will die in a year are p and q respectively. The probability that at the end of the year only of them will be alive, is equal to (Ref. P.K.SHARMA Pg.A8.2 Q.no.11) (A) p + q - pq (B) p + q - (C) p + q – 2pq (D) pq(p + q - pq) Sol: (C) Let ‘A’ denotes the event that dies in a year and ‘B’ denotes the event that dies in a year. Thus, required probability = = P(A) + P(B) – 2(A B) = p + q – 2pq 27. Two fair and ordinary dice are rolled simultaneously 4 times. The probability that both dice will show same outcome exactly twice, is equal to (Ref. P.K.SHARMA Pg.A8.2 Q.no.12) (A) (B) (C) (D) Sol: (A) Probability of showing equal outcomes in any specific roll of dice is , i.e . Thus required probability = Getting exactly 2 successes in 4 trials. = 28. Three numbers are chosen at random with out replacement from the set of integers {1, 2, 3, ……, 10}. The probability that the minimum of the chosen numbers is 3 or the maximum of the chosen numbers is 7, is equal to (Ref. P.K.SHARMA Pg.A8.3 Q.no.14) (A) (B) (C) (D) Sol: (D) Let A : Minimum of the chosen number is 3. B : Maximum of the chosen number is 7. Thus, required probability = 29. Three integers are selected simultaneously from the set of integers {1, 2, 3, 4, ….., 50}. The probability that the selected numbers are consecutive, is equal to (Ref. P.K.SHARMA Pg.A8.3 Q.no.16) (A) (B) (C) (D) None of these Sol: (C) Possible set of three consecutive numbers are {1, 2, 3}, {2, 3, 4}, {3, 4, 5},….{48, 49, 50}. which are 48 in number. Thus, required probability = = 30. Two numbers are selected simultaneously from the set {6, 7, 8, 9, ……, 39}. If the sum of selected numbers is even then the probability that both the selected numbers are odd, is equal to (Ref. P.K.SHARMA Pg.A8.3 Q.no.17) (A) (B) (C) (D) Sol: (D) Total number of even numbers in the set is 18, and total number of odd numbers is 16. A : Sum of selected numbers is even. B : Selected numbers are odd. P(A) = 31. A bag contains 4 red, 5 white and 6 black balls. Three balls are selected from this bag simultaneously. The probability that one of the colour will be missing in the selected balls, is equal to (Ref. P.K.SHARMA Pg.A8.3 Q.no.18) (A) (B) (C) (D) None of these Sol: (A) (i) Selected balls can be W W B or W B B (when red colour is missing). Corresponding probability (ii) Selected balls can be R R W or R W W (when black colour is missing). Corresponding probability (iii) Selected balls can be RBB, RRB (when white colour is missing). Corresponding probability Thus, required probability = 32. Three children are selected at random from a group of 6 boys and 4 girls. It is known that in this group exactly one girl and one boy belong to same parents. The probability that the selected group of children have no blood relations, is equal to (Ref. P.K.SHARMA Pg.A8.3 Q.no.19) (A) (B) (C) (D) Sol: (C) Probability that the boy and his sister both are selected = Thus, required probability = 1 - 33. In a certain city 40% of the people have brown hair, 25% have brown eyes and 15% have brown eyes as well as brown hair. If a person selected at random has brown hair, the probability that the also has brown eyes, is equal to (Ref. P.K.SHARMA Pg.A8.4 Q.no.21) (A) (B) (C) (D) Sol: (D) : Person has brown hair. : Person has brown hair. = 34. A pair of fair and ordinary dice is rolled simultaneously. It is found that they show different outcomes. The probability that the sum of the out comes will be either 6 or 10, is equal to (Ref. P.K.SHARMA Pg.A8.4 Q.no.22) (A) (B) (C) (D) Sol: (A) : Event that dice show different outcomes. : Outcome is either 6 or 10. As 6 can occur in four ways namely (2 , 4), (4 , 2) (1 , 5), (5 , 1) and ten can occur in 2 ways namely (6 , 4) and (4 , 6). Thus, required probability, Exercise # 2 35. In a bag there are 15 red and 5 white balls. Two balls are drawn in succession, without replacement. The first drawn ball is found to be red. The probability that second ball is also red, is equal to (Ref. P.K.SHARMA Pg.A8.5 Q.no.1) (A) (B) (C) (D) Sol: (D) Total number of ways of selecting the second ball Total number of ways of selecting the second ball suitably Thus, required probability . 36. Two fair dice are rolled simultaneously. One of the dice shows four. The probability of other dice showing six, is equal to (Ref. P.K.SHARMA Pg.A8.5 Q.no.2) (A) (B) (C) (D) Sol: (A) Following equally likely outcomes may occur when one of the dice show four; (4, 1), (1, 4) (4, 2), (2, 4) (4, 3), (3, 4) (4, 4), (4, 4) (4, 5), (5, 4) (4, 6), (6, 4). Out of these eleven outcomes exactly 2 outcomes favor the cause of second dice showing a six. Thus, required probability = . 37. Two fair dice are rolled simultaneously. It is found that one of them shows odd prime numbers. The probability that remaining dice also shows an odd prime number, is equal to (A) (B) (C) (D) None of these Sol: (A) 3 and 5 are the only odd prime numbers, among the possible outcomes. The following are the outcomes when one of them show odd prime numbers, (3, 1), (1, 3) (3, 2), (2, 3) (3, 3), (3, 4) (4, 3), (3, 5) (5, 3), (3, 6) (6, 3), (5, 1), (1, 5), (5, 2), (2, 5), (5, 4), (4, 5), (5, 5), (5, 6), (6, 5). Out of these 20 equally likely outcomes exactly 4 favour the presence of odd prime numbers on both dice. Thus, required probability = 38. Let ‘A’, ‘B’ and ‘C’ be three independent events with P(A) = , P(B) = and P(C) = . The probability of exactly 2 of these events occurring, is equal to (A) (B) (C) (D) Sol: (A) Probability of exactly 2 of the three events happening 39. Three different dice are rolled simultaneously, three times. The probability that all of them show different numbers only two times, is equal to (A) (B) (C) (D) Sol: (C) Probability of showing different numbers on a single trial . Thus, required probability = 40. Two subsets A and B of a set S consisting of ‘n’ elements are constructed randomly. The probability that and is equal to (A) (B) (C) (D) Sol: (C) Let ‘A’ has ‘r’ elements, then ‘B’ have all the remaining (n – r) elements and none of the elements that are already present in A. Thus, total number of favourable ways = Hence, required probability = 41. A fair coin is tossed ‘n’ number of times. The probability that head will turn up an even number of times, is equal to (A) (B) (C) (D) Sol: (B) Total outcomes = . Total number of favourable outcomes = = Thus, required probability = 42. A and B each throw a fair dice. The probability that A’s throw is not greater than B’s throw, is equal to (A) (B) (C) (D) Sol: (C) Let ‘A’s throw is r,r = 1, 2, 3, …6, then ‘B’s outcome can be (r, r + 1, ….6), Thus required probability = 43. A and B play a game where each is asked to select a number from 1 to 25. If the numbers selected by A and B match, both of them win a prize. The probability that they win their third prize on 5th game is equal to (A) (B) (C) (D) Sol: (A) Probability of winning the prize in single game = In this case first 4 games, must result in exactly two prizes and 5th game must result in prize. Thus, required probability = 44. A and B are two events such that P(A) = , P(B) = . Then which of the following is not correct? (A) (B) (C) (D) None of these Sol: (D) maximum, {P(A), P(B)} = P(AB’) minimum, {P(A), P(B)} = P(AB) Maxi.{0, 1 –(P(A) + P(B)))} = 0 and P(A B) minimum {P(A), P(B)} = Thus all the statements are correct. 45. A natural number ‘n’ is selected at random from the set of first 100 natural numbers. The probability that is equal to (A) (B) (C) (D) None of these Sol: (C) Thus, favourable number of ways = (47 – 3 + 1) = 45. Thus, required probability = . 46. A bag contains 4 tickets numbered 1, 2, 3, 4 and another bag contains 6 tickets numbered 2, 4, 6, 7, 8, 9. One bag is chosen and a ticket is drawn. The probability that the ticket bears the number 4, is equal to ( (A) (B) (C) (D) Sol: (B) : First bag is chosen, P() = : Second bag is chosen, P() = A : Drawn number is 4. Now, P(A) = P() . P(A/) + P() . P(A/) 47. A bag contains four tickets marked with numbers 112, 121, 211, 222. One ticket is drawn at random from the bag. Let (i = 1, 2, 3) denote the event that digit on the selected ticket is 2, then which of the following is not correct? (A) and are independent (B) and are independent (C) and are independent (D) ,, are independent Sol: (D) and are independent. and are independent. and are independent. and are not independent. 48. A real estate man has eight master keys to open several new homes. Only one master key will open any given home. If 40% of these homes are usually left unlocked, the probability that the real estate man can get into a specific home, if it is given that he selected 3 keys randomly before leaving his office, is equal to (A) (B) (C) (D) Sol: (A) : Specific home is locked. : Specific home is unlocked. A : Real estate man get in to the home. P(A) = P() . P(A/) + P() . P(A/) 49. Two numbers and are chosen at random (with out replacement) from the set {1, 2, 3, …., 5n}. the probability that is divisible by 5, is equal to (A) (B) (C) (D) Sol: (C) We can segregate the numbers as multiple of five (5 ), 5 + 1, 5 + 2, 5 + 3, 5 + 4, as follows: We can select either both the number from , or any two numbers from the first four rows. (As square of any number that is not a multiple of three will be in the form of 5 ’ + 1 or 5 ’ – 1) Thus, required probability Exercise # 3 50. Two persons are selected at random from n persons seated in a row . The probability that the selected persons are not seated consecutively, is equal to (A) (B) (C) (D) Sol: (A) Total ways of selecting two persons = = Total ways of selecting two consecutively seated persons = = (n – 1) Thus, required probability = = 51. Three numbers are selected simultaneously from the set {1, 2, 3, …., 25}. The probability that the product of selected numbers is divisible by 4, is equal to (A) (B) (C) (D) Sol: (C) Product can’t be divisibly by 4 in the following mutually exclusive cases; (i) All the selected numbers are odd. corresponding ways = 286 (ii) Two of the selected numbers are odd and another is a multiple of 2 only. Corresponding ways = = 468 Thus, required probability = 1 - 52. A man alternatively tosses a fair coin and rolls a fair ordinary dice. He starts with the coin. The probability that he gets a tail on the coin before getting 5 or 6 on the dice, is equal to (A) (B) (C) (D) Sol: (A) Probability of getting 5 or 6 on a specific roll of dice and, probability of getting a tail = . The desired outcome can happen, in general, on (2r + 1 )th trial. That means first 2r trials should result neither in tail nor in 5 or 6, and (2r + 1 )th trial must result in tail. It the conesponding probability is then Thus, required probability . 53. Three ordinary and fair dice are rolled simultaneously. The probability of the sum of outcomes being atleast equal to 8, is equal to (A) (B) (C) (D) Sol: (D) Let the outcome on dice be and . we first try to find the number of ways in which sum of outcomes is at the most equal to seven. For this, , where . i.e., where are non-negative integers. This can happen in i.e., ways. Thus, required probability 54. Two squares are chosen from the squares of a ordinary chess board. It is given that the selected squares do not belong to the same row or column. The probability that the chose squares are of same colour, is equal to (A) (B) (C) (D) Sol: (A) Let the first chosen square is white. In this case total number of squares that don’t belong the row or column associated with the selected square is 49. Out of these 25 are white and 24 are black. Thus, corresponding probability = Similarly, if the first square is black, then the corresponding probability = Thus required probability . 55. A committee consists of 9 experts taken from three institutions A, B and C, of which 2 are from A, 3 from B and 4 from C. If three experts resign from the committee, then the probability of exactly two of the resigned experts being from the same institution, is equal to (A) (B) (C) (D) Sol: (C) These are three mutually exclusive cases: (i) 2 are from A and another is from either from B or C. Corresponding probability . (ii) 2 are from B and another is either from A or C. Corresponding probability . (iii) 2 are from C and another is either from A or B. Corresponding probability . Thus, required probability = = . 56. The sum of two natural numbers and is known to be equal to 100. The probability that their product being greater than 1600, is equal to (A) (B) (C) (D) Sol: (D) Total number of ways in which = 100 is equal to 99. Now, Thus, number of favorable ways = 79 – 21 + 1 = 59 Hence, required probability = . 57. A person while dialing a telephone number, forgets the last three digits of the number but remembers that exactly two of them are same. He dials the number randomly. The probability that he dialed the correct number, is equal to. (A) (B) (C) (D) Sol: (D) Total number of ways of dialing the last three digits such that exactly 2 of them are same Thus, required probability = . 58. A bag has 3 white balls and 2 red balls. Another bag has 4 white and 6 red balls. A ball is drawn randomly from bag and with out seeing it’s colour, is being put in bag . Now a ball is drawn from bag . The probability of both the drawn balls, of being same colour, is (A) (B) (C) (D) None of these Sol: (C) : Ball drawn from is white. : Ball drawn from is red. A : Both the drawn balls are of same colour. P(A) = P().P(A/) + P().P(A/) = 59. A bag contains ‘W’ white balls and ‘R’ red balls. Two players and alternately draw a ball from the bag, replacing the ball each time after the draw, till one of them draws a white ball and wins the game. beings the game. The probability of being the winner, is equal to (A) (B) (C) (D) Sol: (D) Probability of drawing a white ball at any draw = . Now, can be the winner in general on , r 1, draw. That means in the first (2r – 1) draws no player draws a white ball and draw results in a white ball for . If the corresponding probability is , then = Hence, required probability = = 60. If the papers of 4 students can be checked by any one of the 7 teachers, then the probability that all the 4 papers are checked by exactly 2 teachers, is equal to (Ref. (A) (B) (C) (D) Sol: (B) Total ways in which papers can be checked is equal to . Now two teachers who have to check all the papers can be selected in ways and papers can be checked by them in favourable ways. Thus, required probability = 61. A bag contains 100 balls numbered from 1 to 100. Four balls are removed at random with out replacement. The probability that the number on the last ball is smaller than the number on first ball, is equal to ( (A) (B) (C) (D) Sol: (D) If and be the numbers on the drawn balls (in that order) than, we must have . Total number of order relations between and is 4! i.e., 24. Now total number of favourable order relations have to counted. (i) If ‘’ is the largest, then favorable order relations are 3! = 6. (ii) If is the second largest, then favourable order relations are 2(2!) = 4. (iii) In is the third largest, then favourable order relations are 2! = 2. Thus, required probability = Alternate solution Clearly and are unequal. Thus there are only two cases or . Thus, required probability = 62. The probability that an archer hits the target when it is windy is equal to 2/5, when it is not windy his probability of hitting the target is 7/10. On any shot the probability of gust of wind is . The probability that there is no gust of wind on the occasion when he missed the target, is equal to (A) (B) (C) (D) Sol: (C) : There is a gust of wind. : There is no gust of wind. A : archer misses the target. P(A) = P().P(A/) + P().P(A/) Now, required probability = 63. Three numbers are selected from the set {1, 2, 3, …., 23, 24}, with out replacement. The probability that the formed numbers from an A.P. is equal to (A) (B) (C) (D) None of these Sol: (C) Let the selected numbers be and . We must have . Thus must be even. That means and both must have same nature. Thus, required probability = .