Comprehensive CBSE Question Bank in Mathematics-XII (Term-II)

Three Dimensional Geometry 309 (2, 3, 1) B(3, –1, 1) A(1, 0, 2) C(1, 2, 1) Based on the above information, answer the following questions: (i) The equation of the plane passing through the points A, B and C is (a) 3x – 2y + 4z = –11 (b) 3x + 2y + 4z = 11 (c) 3x – 2y – 4z = 11 (d) –3x + 2y + 4z = –11 (ii) The height of the tower from the ground is (a) 5 units (b) 7 units (c) 6 units (d) 8 units 29 29 29 29 (iii) The equation of the perpendicular line drawn from the top of the tower to the ground is (a) x − 1 = y + 3 = z−5 (b) x−2 = y − 3 = z−1 2 1 −2 −3 2 −4 (c) x − 2 = y − 3 = z − 1 (d) x+1 = y+3 = z − 5 3 2 4 −2 −1 2 (iv) The coordinates of the foot of the perpendicular drawn from the top of the tower to the ground are (a) ⎛ 43 , −77 , −9 ⎞ (b) ⎛ 9 , −1 , −10 ⎞ ⎜⎝ 29 29 29 ⎟⎠ ⎝⎜ 7 7 7 ⎠⎟ (c) ⎛ −43 , 77 , −9 ⎞ (d) ⎛⎝⎜ 43 , 77 , 9 ⎟⎞⎠ ⎜⎝ 29 29 29 ⎟⎠ 29 29 29 (v) The area of Δ ABC is (a) 29 sq. units (b) 29 sq. units 4 2 (c) 39 sq. units (d) 39 sq.units 2 4

310 Mathematics—XII Answers 1. (i) (b) (ii) (d) (iii) (c) (iv) (c) (v) (b) 2. (i) (c) (ii) (b) (iii) (b) (iv) (c) (v) (a) 3. (i) (c) (ii) (c) (iii) (a) (iv) (c) 4. (i) (c) (ii) (b) (iii) (b) (iv) (c) (v) (b) 5. (i) (c) (ii) (b) (iii) (d) (iv) (b) (v) (a) 6. (i) (c) (ii) (d) (iii) (c) (iv) (b) (v) (d) 7. (i) (a) (ii) (b) (iii) (c) (iv) (d) (v) (a) 8. (i) (c) (ii) (b) (iii) (d) (iv) (a) (v) (b) 9. (i) (a) (ii) (c) (iii) (b) (iv) (d) (v) (b) 10. (i) (d) (ii) (a) (iii) (c) (iv) (d) 11. (i) (b) (ii) (a) (iii) (c) (iv) (d) ❏❏❏

Probability 311 UNIT VI: PROBABILITY CHAPTER 6: Probability Syllabus: Conditional probability, mutliplication theorem on probability, independent events, total probability, Bayes’ theorem, Random variable and its probability distribution. IMPORTANT RESULTS AND FORMULAE 1. RECAPITULATION (i) Sample space: The set of all possible outcomes of a random experiment is called the sample space. For example 1. In tossing a coin, the sample space S = {H, T} ⇒ n(S) = 2. 2. In tossing a coin twice, S = {HH, HT, TH, TT} ⇒ n(S) = 4. 3. In throwing a die twice, S = {(1, 1), (1, 2), ..... (1, 6), (2, 1), ....(6, 6)} ⇒ n(S) = 36. (ii) Event: A subset of a sample space is called an event. For example 1. In tossing a coin once, two events are“getting head” and “getting tail”. Symbolically we can write it as A = {H} and B = {T}. 2. In throwing a die once, the event of getting an odd number is A = {1, 3, 5}, the event of getting even number is B = {2, 4, 6}, the event of getting prime number is C = {2, 3, 5} and so on. (a) Elementary event or simple event: If a random experiment is performed each of its outcomes are called elementary events or simple events. For example, in tossing a coin “getting head” is a simple event. (b) Compound event: An event associated with a random experiment is a compound event if it is the union of one or more simple events. For example, in throwing die twice, the event “getting sum 8” is a compound event. i.e., A = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)} is a compound event. (c) Sure or certain event: An event which is sure to occur is called a sure event. For example, in throwing a die the event of “getting a number less than 7” is a sure event. (d) Impossible event (null event): An event which can never occur is called an impossible event. For example, in throwing a die, the event of “getting a number greater than 6” is an impossible event. (e) Mutually exclusive events: Two or more events of a random experiment are said to be mutually exclusive events if the occurrence of any one of the events prevents the occurrence of the other event. i.e., two events A and B of a random experiment are mutually exclusive events if A ∩ B = φ. (f) Exhaustive events: Two or more events associated with a random experiment are exhaustive events if their union is the sample space. 311

312 Mathematics—XII (iii) Probability of an event: If there are n elementary events of a random experiment and m of them are favourable to the event A, then the probability of happening or occurring the event A is m . n Thus, P(A) = m . n i.e., P(A) = No. of outcomes favourable to A . Total possible outcomes Remarks 1. P(A) = 1 if A is the sure event and P(A) = 0 if A is the impossible event. i.e., P(S) = 1, P(φ) = 0 and so, 0 ≤ P(A) ≤ 1. 2. If A denote the event of “not happening A”, then P(A) = 1 – P (A) i.e. P(not happening an event A) = 1 – P(happening the event A). (iv) Addition theorem on Probability: (a) If A and B are two events associated with a random experiment, then P(A ∪ B) = P(A) + P(B) – P(A ∩ B). (b) If A and B are two mutually exclusive events, then P(A ∪ B) = P(A) + P(B). (c) If A, B, C are any three events associated with a random experiment, then P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(A ∩ C) + P(A ∩ B ∩ C). (d) If A, B, C are three mutually exclusive events, then P(A ∪ B ∪ C) = P(A) + P(B) + P(C). (e) For any two events A and B, (i) Probability of occurrence of A only = P(A ∩ B) = P(A) − P(A ∩ B) . (ii) Probability of occurrence of B only = P(A ∩ B) = P(B) − P(A ∩ B) . (iii) Probability of occurrence of exactly one of A and B = P(A ∩ B) + P(A ∩ B) = P(A) + P(B) − 2P(A ∩ B) = P(A ∪ B) − P(A ∩ B) . 2. CONDITIONAL PROBABILITY Let A and B be two events associated with a random experiment. Then the probability of the occurrence of the event A under the condition that B has already happened, P(B) ≠ 0, is the conditional probability and is denoted by P(A/B). i.e., P(A/B) is the probability of the event A given that B has already happened. Similarly, P(B/A), P(A) ≠ 0 is defined as the probability of the event B when A has already happened. The symbols P(A/B) has the following meaning also. P(A/B) = probability of the occurrence of A when B occurs. OR Probability of the occurrence of A when B is taken as the sample space. OR Probability of the occurrence of A with respect to B. Standard Results on Conditional Probability: If A and B are any two events of a random experiment, then (i) P(A/B) = P(A ∩ B) , P(B) ≠ 0 P(B)

Probability 313 (ii) P(B/A) = P(A ∩ B) , P(A) ≠ 0 . P(A) In practical problems, it is easier to use the formulae (i) P(A/B) = No. of outcomes favourable to A ∩ B = n(A ∩ B) . no. of outcomes favourable to B n(B) (ii) P(B/A) = No. of outcomes favourable to A ∩ B = n(A ∩ B) . no. of outcomes favourable to A n(A) Properties of Conditional Probability (i) If A and B are any two events associated with a sample space S of a random experiment, then 0 ≤ P(A/B) ≤ 1. (ii) If A is an event associated with the sample space S of a random experiment, then P(S/A) = P(A/A) = 1. (iii) If A and B are two events associated with the sample space S of a random experiment and C is an event such that P(C) ≠ 0, then P((A ∪ B)/C) = P(A/C) + P(B/C) – P((A ∩ B)/C). (iv) If A and B are any two events associated with a random experiment, then P(A/B) = 1 – P(A/B). Example 1. A pair of dice is thrown. Find the probability that the sum is 10 or more if Solution. 5 appears in the first die. Let A be the event of getting sum 10 or more and B the event that 5 appears in the first die. Then the required probability = P(A/B) = P(A ∩ B) . P(B) Now A = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)} B = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} So, we get A ∩ B = {(5, 5), (5, 6)}. Hence the required probability = P(A/B) = P(A ∩ B) = 2 / 6 = 2 = 1 . P(B) 36 36 6 3 Alternatively, the required probability = n(A ∩ B) = 2 = 1 . n(B) 6 3 Example 2. In a certain school, 20% of the students failed in English, 15% failed in Mathematics and 10% failed in both English and Mathematics. A student is selected at random. If he failed in English, what is the probability that he also failed in Mathematics? Solution. Let E and M are the events that a student fails in English and Mathematics. Example 3. Given, P(E) = 20 = 1 , P(M) = 15 = 3 100 5 100 20 10 1 and P(E ∩ M) = 100 = 10 . Required probability = P(M/E ) = P(M ∩ E) = 1 / 10 = 1 . P(E) 1/5 2 Two integers are selected from 1 to 11. If the sum is even, find the probability that both the numbers are odd. Solution. Out of 11 numbers, 5 are even and 6 are odd. Let A be the event that both are odd and B be the event that sum is even.

314 Mathematics—XII So, the required probability P (A/B) = Probability that the two numbers chosen are odd given that the sum of the numbers is even = 6 C2 = 15 = 3 . 5 C2 + 6C2 10 + 15 5 Example 4. If A and B are two events such that P(A) = 0.5, P(B) = 0.6 and P(A ∪ B) = 0.8, Solution. find P(A/B) and P(B/A). We have P(A ∪ B) = P(A) + P(B) – (A ∩ B) ⇒ P(A ∩ B) = 0.5 + 0.6 – 0.8 = 0.3. ∴ P(A/B) = P(A ∩ B) = 0.3 = 1 . P(B) 0.6 2 Also, P(B/A) = P(A ∩ B) = 0.3 = 3 . P(A) 0.5 5 3. The probability of both the events A and B If A and B are any two events of a random experiment, then the probability of both the events A and B is denoted by P(A ∩ B) or simply P(AB). 4. Multiplication Theorem on Probability If A and B are any two events of a random experiment, then P(A ∩ B) = P(A) P(B/A) if P(A) ≠ 0 and P(A ∩ B) = P(B) P(A/B) if P(B) ≠ 0. This theorem can be extended further. i.e., if A, B, C are three events of a random experiment, then P(A ∩ B ∩ C) = P(A) P(B/A) P(C/A ∩ B). Example 5. A bag contains 10 white and 15 black balls. Two balls are drawn in succession without replacement. What is the probability that first is white and second is black? Solution. Let W = getting white ball in the first draw and B = getting black ball in the second draw. The required probability = P(W ∩ B) = P(W) P(B/W). Now, P(W) = 10 = 2 . 25 5 15 5 P(B/W) = prob. of drawing a black ball after drawing white = 24 = 8 (Now there are only 24 balls out of which 15 black balls) Hence, the required probability = 2 × 5 = 1 . 5 8 4 Example 6. A bag contains 19 tickets, numbered from 1 to 19. A ticket is drawn and then another ticket is drawn without replacement. Find the probability that both tickets will show the even numbers. Solution. Let A be the event of drawing an even numbered ticket in the first draw and B be the event of getting even number in the second draw. There are 9 even numbered tickets out of 19 tickets. Clearly, P(A) = 9 and P(B/ A) = 8 = 4 . 19 18 9 Hence, the required probability = P(A ∩ B) = P(A) P(B/A) = 9 × 4 = 4 . 19 9 19

Probability 315 Example 7. If A and B are two events such that P(A) = 0.3, P(B) = 0.6 and P(B/A) = 0.5, find P(A/B) and P(A ∪ B). Solution. We have P(A ∩ B) = P(A)·P(B/A) = 0.3 × 0.5 = 0.15. ∴ P(A/B) = P(A ∩ B) = 0.15 = 1 . Also, P(B) 0.6 4 P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.3 + 0.6 – 0.15 = 0.75. 5. Theorem on total probability. Let E1, E2, E3, ......, En are n mutually exclusive and exhaustive events of a random experiment. If A is any event which occurs with E1 or E2 or E3 or ......or En, then, P(A) = P(E1).P(A/E1) + P(E2).P(A/E2) + ...... + P(En).P(A/En). Example 8. An urn A contains 2 white and 4 black balls. Another urn B contains 5 white and 7 black balls. A ball is transferred from urn A to B. Then a ball is drawn from urn B. Find the probability that it is a white ball. Solution. There are two cases (i) Transferring white ball from urn A to B and then drawing white ball from B. (ii) Transferring black ball from urn A to B and then drawing white ball from B. Let E1 = transferring white E2 = transferring black W = drawing white Now, P(E1) = 2 = 1 , P(E2 ) = 4 = 2 . 6 3 6 3 P(W/E1) = P (drawing white after transferring white) = 6 . 13 (After transferring white urn B will have 6 white out of 13 balls) P(W/E2) = P (drawing white after transferring black) = 5 . 13 (After transferring black urn B will have 5 white out of 13 balls) Required probability = P(W) = P(E1)⋅P(W/E1) + P(E2)⋅P(W/E2) = 1 × 6 + 2 × 5 = 16 . 3 13 3 13 39 6. Independent events. Two events A and B are said to be independent, if the occurrence or non-occurrence of one of the events does not affect the probability of the occurrence or non-occurrence of the other event. i.e., if A and B are independent events associated with a random experiment, then P(A/B) = P(A), P(B/A) = P(B) and so P(A ∩ B) = P(A) P(B). Theorem 1: Two events A and B associated with a random experiment are independent events, if and only if P(A ∩ B) = P(A)P(B). This can be extended further. Theorem 2: If A1, A2, A3, ......,An are independent events associated with a random experiment, then P(A1 ∩ A2 ∩ A3 ...... ∩ An) = P(A1) P(A2) P(A3) ...... P(An). 7. Pair wise independent events: The events A1, A2, A3, ......,An associated with a random is experiment are said to be pair wise independent event if P(Ai ∩ Aj ) = P(Ai) P(Aj) for all i ≠ j; i, j = 1, 2, 3, ......, n.

316 Mathematics—XII 8. Mutually independent seivmeunlttsa:nTehoeuesvoecnctusrAre1,nAce2,oAf3a, n...y...f,iAnintearneummubteuraollfytihnedmepisenedquenatl if the probability of the to the product of their separate probabilities. i.e., the events AA31,..A...2., ∩A3A, .n..)..=., AP(nAo1f)aPr(aAn2)dPo(mA3e)x.p...e..riPm(Aenn)t. are mutually independent, if P(A1 ∩ A2 ∩ Remark: It follows from the definition that the mutually independent events are always pair wise independent but the converse need not be true. Theorem 3: If A and B are independent events associated with a random experiment, then (i) A and B are independent i.e., P(A ∩ B) = P(A).P(B) = [1 − P(A)]P(B) . (ii) A and B are independent i.e., P(A ∩ B) = P(A).P(B) = P(A)[1 − P(B)] . (iii) A and B are independent i.e., P(A ∩ B) = P(A).P(B) = [1 − P(A)][1 − P(B)] . 9. The probability of the occurrence of at least one of the events A and B If A and B are two independent events then the probability of the occurrence of at least one of the events A and B = P(A ∪ B) = 1 – P(A)P(B) . This can be extended further. ip.er.o, bIaf bAil1i,tyAo2,f Ath3e, .o..c..c.u, rArnenacree independent events of a random experiment, then the of at least one of the events is P(A1 ∪ A2 ∪ A3 ∪ ...... ∪ An) = 1 − P(A1)P(A2 )P(A3 ) ...... P(An ) . i.e., P(occurrence of at least one of the events) = 1 – P(occurrence of none of the events) Example 9. A die is thrown once. If A is the event “the number appearing is a multiple of 3” and B is the event “the number appearing is even”. Are the events independent? Solution. A = {3, 6} and B = {2, 4, 6}. 2 1 Then P(A) = 6 = 3 and P(B) = 3 = 1 . Also P(A ∩ B) = 1 . 6 2 6 Clearly, P(A ∩ B) = P(A) P(B). So, A and B are independent. Example 10. If A and B are two events such that P(A) = 1 , P(B) = 1 and P(A ∩ B) = 1 , 4 2 8 find P(not A and not B). Solution. Since P(A ∩ B) = P(A)P(B), A and B are independent. Now, P(not A and not B) = P(A ∩ B) = P(A)P(B) (since A and B are independent) = ⎛ 1 − 1 ⎞ ⎛ 1 − 1⎞ = 3 × 1 = 3 . ⎝⎜ 4 ⎟⎠ ⎝⎜ 2 ⎠⎟ 4 2 8 Example 11. If P(A) = 0.4, P(B) = p, P(A ∪ B) = 0.6, A and B are independent, find the value of p. Solution. Since A and B are independent events, P(A ∩ B) = P(A)P(B). We have, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) ⇒ P(A ∪ B) = P(A) + P(B) – P(A)P(B) (Since A and B are independent.) ⇒ 0.6 = 0.4 + p – 0.4p ⇒ 0.6p = 0.2 ⇒ p= 1 . 3 Example 12. Two dice are thrown. Find the probability of getting an odd number on the first die and a multiple of 3 on the other. Solution. Let A = getting an odd number on the first die

Probability 317 B = getting a multiple of 3 on the second die 1 1 3 Then A = {1, 3, 5}, B = {3, 6} ⇒ P(A) = 2 , P(B) = Clearly, A and B are independent. 1 1 1 2 3 6 Hence, the required probability = P(A ∩ B) = P(A)P(B) = × = . 10. Probability of neither A nor B: If A and B are any two events of a random experiment, then P(neither A nor B) = P(A ∩ B) (i) If A and B are independent events, then P(neither A nor B) = P(A ∩ B) = P(A)P(B) . (ii) If A and B are any two events, then P(neither A nor B) = P(A ∩ B) = P(A ∪ B) = 1 − P(A ∪ B) = 1 − P(A or B) . 11. Probability of exactly one of the events: If A and B are any two independent events of a random experiment, then P(exactly A) = P(A ∩ B) = P(A) ⋅ P(B) . Example 13. A speaks truth in 60% of the cases and B in 90% of the cases. In what percentage of the cases they are likely to contradict each other in stating the same fact? Solution. Let E = A speaks truth F = B speaks truth ⇒ P(E) = 60% = 6 , P(F) = 90% = 9 10 10 Given, they contradict each other in stating the same fact. Then we have the following cases. (i) A speaks truth and B tell lie i.e., E ∩ F (ii) A tell lie and B speaks truth i.e., E ∩ F . So, the required probability = P(E ∩ F) + P(E ∩ F) . = P(E) ⋅ P(F) + P(E) ⋅ P(F) . = 6 ⎛ 1 − 9 ⎞ + ⎛ 1 − 6 ⎞ 9 = 42 = 21 . 10 ⎜⎝ 10 ⎟⎠ ⎜⎝ 10 ⎟⎠ 10 100 50 Example 14. A problem in Mathematics is given to three students whose probabilities of solving it are 1 , 1 , 1 respectively. What is the probability that the problem is solved? 2 3 4 Solution. Let A, B, C be the respective events of solving the problem. 1 1 1 Then, P(A) = 2 , P(B) = 3 and P(C) = 4 . Given the problem is solved i.e., at least one of them solves. ∴ Required probability = 1 – P(none of them solved) = 1 – P (A ∩ B ∩ C) = 1 − P(A)P(B)P(C) = 1 − ⎛ 1 − 1⎞ ⎛⎜⎝ 1 − 1⎞ ⎛ 1 − 1⎞ ⎝⎜ 2 ⎠⎟ 3 ⎟⎠ ⎜⎝ 4 ⎟⎠ = 1 − 1 × 2 × 3 = 1 − 1 = 3 . 2 3 4 4 4 12. BAYE’S THEOREM: Let S be the sample space and let E1, E2, E3, ......, En are n mutually exclusive and exhaustive events of a random experiment. If A is any event of non-zero probability which occurs with E1 or E2 or ...... or En, then

318 Mathematics—XII P(Ei/A) = P(Ei ).P(A / Ei ) for any i = 1, 2, 3, ......, n. n ∑ P(Ei ).P(A / Ei ) i =1 Remark: The following terminologies are usually used when Baye’s theorem is applied. The events E1, E2, ......, En are called hypotheses. The probabilities P(E1), P(E2), ......, P(En) are called priori probabilities as they exist before any information from the experiment. The probabilities P(Ei/A), i = 1, 2, ......, n are called posterioiri probabilities of the hypothesis Ei. STEPS FOR CALCULATION 1. Identify mutually exclusive and exhaustive events E1, E2, ......, En. 2. Find P(E1), P(E2) ......, P(En) and check that the sum of the probabilities is 1. 3. Identify the event A and calculate P(A/E1), P(A/E2), ......, P(A/En). 4. Identify the required event Ei/A and calculate P(Ei/A) using Baye’s theorem. Example 15. A bag X contains 2 white and 3 red balls and a bag Y contains 4 white and 5 red balls. One ball is drawn at random from one bag and found to be red. Find the probability it was drawn from the bag Y. Solution. Let EEA12 = the bag X is chosen = the bag Y is chosen = the ball is red. We have to find Pit(iEs2/reAd)) (i.e., the probability that the ball drawn from bag Y, given that We have by Baye’s theorem, P(E2/A) = P(E1 P(E2 ) ⋅ P(A/ E2 ) A / E2 ) . ).P(A/ E1) + P(E2 ).P( Since, the bags are equally likely to be selected, P(E1) = P(E2) = 1 . 2 Also, P(A/E1) = P(drawing red ball from bag X) = 3 . 5 P(A/E2) = P(drawing red ball from bag Y) = 5 . 9 1 × 5 5 5 45 25 2 9 9 9 52 52 ∴ P(E2/A) = 1 3 1 5 = 3 5 = × = . 2 5 2 9 5 + 9 × + × 13. Random variable: In practical applications of probability, it will be more convenient to work with numbers than with outcomes such as “heads”, “sixes”, “boys”, etc. A random variable is often described as a variable whose values are determined by the outcomes of the random experiment. A random variable is usually denoted by X. A random variable can take any real value. In other words, a random variable is a real valued function whose domain is the sample space associated with a random experiment. Illustration: Consider the random experiment of tossing a coin twice. Then the sample space is S = {HH, HT, TH, TT}.

Probability 319 If we define X = the number of heads, then X can assume the values 0 for TT, 1 for HT or TH, 2 for HH. 14. Discrete random variable: Discrete random variable is a random variable which can take only finite number of values. 15. Probability distribution of a random variable: The tabular description of values of the random variable X and the corresponding probabilities is known as probability distribution. Definition: If a random variable X takes the valuesx1, x2, x3, ......, xn with corresponding probabilities p1, p2, p3, ......,pn, then X: x1 x2 x3 ..... xn P(x): p1 p2 p3 ..... pn where pi > 0 and ∑ pi = 1 , is the probability distribution of X. Example 16. Find the probability distribution of X, the number of heads, when a coin is tossed twice. Solution. When a coin is tossed twice, there may be one head, two heads or no head. Thus, X = 0, 1, 2. Corresponding to no head, one head or two heads. Now, P(X = 0) = P(getting no head) = P(TT) = 1 . 4 P(X = 1) = P(getting one head) = P(HT or TH) = 2 = 1. 4 2 1 P(X = 2) = P(getting two heads) = P(HH) = 4 . Hence, the probability distribution of X, the number of heads is X: 0 1 2 P(X) : 1 1 1 4 2 4 MISCELLANEOUS EXAMPLES Example 1. Three events A, B and C have probabilities 2 , 1 and 1 respectively given that Solution. 5 3 2 P(A ∩ C) = 1 and P(B ∩ C) = 1 , find the values of P(C/B) and P(A′ ∩ C′). 5 4 [NCERT (EP)] We have P(A) = 2 , P(B) = 1 , P(C) = 1 , P(A ∩ C) = 1 and P(B ∩ C) = 1 . 5 3 2 5 4 P(C/B) = P(C ∩ B) = P(B ∩ C) = 1/4 = 3 . P(B) P(B) 1/3 4 Also, P(A′ ∩ C′) = P((A ∪ C)′) = 1 – P(A ∪ C) = 1 – [P(A) + P(C) – P(A ∩ C)] = 1 – ⎡2 + 1 − 1⎤ = 3 . ⎣⎢ 5 2 5 ⎦⎥ 10

320 Mathematics—XII Example 2. A coin is tossed twice and the four possible outcomes are assumed to be equally Solution. likely. If E is the event: “both head and tail have occurred”, and F the event: “at most one tail is observed”, find P(E), P(F), P(E/F) and P(F/E). Here S = {HH, HT, TH, TT}. We have E = {HT, TH} and F = {HH, HT, TH}. ∴ E ∩ F = {HT, TH} ∴ P(E) = n (E) = 2 = 1 P(F) = n (E) = 3 n (S) 4 2 n (S) 4 P(E/F) = n (E ∩ F) = 2 P(F/E) = n (F ∩ E) = n (E ∩ F) = 2 = 1. n (F) 3 n (E) n (E) 2 Example 3. A die is thrown twice and the sum of the number appearing is observed to be 6. What Solution. is the conditional probability that the number 4 has appeared at least once? (NCERT) Let S be the sample space of the experiment. ∴ S = {(1, 1), (1, 2), ......, (6, 5), (6, 6)}. Let A = event of getting sum 6 and B = event of getting 4 at least once. ∴ A = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)} and B = {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (1, 4), (2, 4), (3, 4), (5, 4), (6, 4)}. ∴ P(A) = 5 and P(B) = 11 36 36 Also, A ∩ B = {(4, 2), (2, 4)} ∴ P(A ∩ B) = 2 Now, required probability 36 = probability of getting 4 on at least one die given that sum is 6 = P(B/A) = P(B ∩ A) = P(A ∩ B) = 2/36 = 2 . P(A) P(A) 5/36 5 Alternatively, P(B/A) = Number of cases favourable to ‘A ∩ B’ = 2 . Number of cases favourable to A 5 Remark. In practical problems, it would be easier to use the formula: P(A/B) = Number of cases favourable to ‘A ∩ B’ Number of cases favourable to B Example 4. One card is drawn from a well-shuffled pack of 52 cards. If E is the event “the Solution. card drawn is either a king or an ace” and F is the event “the card drawn is either an ace or a jack”, then find the probability of the conditional event E/F. There are 4 kings and 4 aces in the pack. ∴ P(E) = 4+4 = 2 52 13 There are 4 aces and 4 jacks in the pack. ∴ P(F) = 4+4 = 2 52 13

Probability 321 The event E ∩ F contain 4 aces. ∴ P(E ∩ F) = 4 = 1 52 13 ∴ Required probability = P(E/F) = P(E ∩ F) = 1/13 = 1 . P(F) 2/13 2 Example 5. A couple has 2 children. Find the probability that both children are boys, if it is Solution. known that (i) at least one of the children is a boy (ii) at least one of the children is a girl (iii) the older child is a boy (iv) the older child is a girl. Example 6. Solution. (NCERT; CBSE 2014 C; 2018 SP) Here S = {B1B2, G1G2, G1B2, B1G2} The suffixes ‘1’ and ‘2’s refer to older child and younger child, respectively. Let A = event that both children are boys. ∴ A = {B1B2} (i) Let B = event that at least one of the children is a boy. ∴ B = {B1B2, G1B2, B1G2} ∴ A ∩ B = {B1B2} ∴ Required probability = P(A/B) = n(A ∩ B) = 1 . n(B) 3 (ii) Let B = event that at least one of the children is a girl. ∴ B = {G1G2, G1B2, B1G2} ∴ A∩B=φ ∴ Required probability = P(A/B) = n(A ∩ B) = 0 = 0. n(B) 3 (iii) Let B = event that the older child is a boy. ∴ B = {B1B2, B1G2} ∴ A ∩ B = {B1B2} ∴ Required probability = P(A/B) = n(A ∩ B) = 1 . n(B) 2 (iv) Let B = event that the older child is a girl. ∴ B = {G1B2, G1G2} ∴ A∩B=φ ∴ Required probability = P(A/B) = n(A ∩ B) = 0 = 0. n(B) 2 A committee of 4 students is selected at random from a group consisting 8 boys and 4 girls. Given that there is at least one girl in the committee, calculate the probability that there are exactly 2 girls in the committee. [NCERT (EP)] Let A = event that there is at least one girl in the committee and B = event that there are exactly 2 girls in the committee. ∴ Required probability = P(B/A) = P(B ∩ A) P(A) P(A) = 1 – (No girl is selected) =1– 8 C4 =1– 8 C4 =1– 70 =1– 14 = 85 8 + 4 C4 12 C4 495 99 99

322 Mathematics—XII P(B ∩ A) = P(2 girls are selected) = 8 C2 × 4C2 = 28 × 6 = 56 8 + 4 C4 495 165 ∴ Required probability = P(B/A) = P(B ∩ A) = 56/165 = 56 × 99 = 168 . P(A) 85/99 165 × 85 425 Example 7. Three fair coins are tossed. Find the probability that they are all tails, if: Solution. (i) at least one of the coins show tail. (ii) two coins show tail. Example 8. Solution. (iii) at least two coins show head. (iv) at most one coin show head. Here S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} Let A = event of getting all tails. ∴ A = {TTT} (i) Let B = event that at least one of the coins show tail. ∴ B = {HHT, HTH, THH, HTT, THT, TTH, TTT} ∴ A ∩ B = {TTT} ∴ Required probability = P(A/B) = n(A ∩ B) = 1 . n(B) 7 (ii) Let B = event that two coins show tail. ∴ B = {HTT, THT, TTH} ∴ A ∩ B = φ ∴ Required probability = P(A/B) = n(A ∩ B) = 0 = 0. n(B) 3 (iii) Let B = event that at least two coins show head. ∴ B = {HHH, HHT, HTH, THH} ∴ A ∩ B = φ ∴ Required probability = P(A/B) = n(A ∩ B) = 0 = 0. n(B) 4 (iv) Let B = event that at most one coin show head. ∴ B = {HTT, THT, TTH, TTT} ∴ A ∩ B = {TTT} ∴ Required probability = P(A/B) = n(A ∩ B) = 1 . n(B) 4 Consider the experiment of throwing a die. If a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one throw of die shows a 3’. (NCERT) Let S be the sample space of the experiment. ∴ S = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6), 1H, 1T, 2H, 2T, 4H, 4T, 5H, 5T} The outcomes of S are not equally likely. In throwing a die, the probability of getting a multiple of 3 is 2 i.e., 1 and not getting a multiple of 3 is 6 3 1– 1 = 2 . The first 12 outcomes in S are equally likely and sum of their 3 3

Probability 323 probabilities is 1 . Thus, the probabilities of each of first 12 outcomes is 3 1 ⎛ 1 ⎞ = 1 . The remaining 8 outcomes in S are also equally likely and 12 ⎜⎝ 3 ⎠⎟ 36 sum of their probabilities is 2 . Thus, the probabilities of each of remaining 3 8 outcomes is 1 ⎛ 2 ⎞ = 1 . 8 ⎜⎝ 3 ⎠⎟ 12 Let A = event of getting tail on the coin and B = event of getting at least one 3 ∴ A = {1T, 2T, 4T, 5T} and B = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 3)} ∴ A ∩ B = φ ∴ P(A ∩ B) = 0 Also, P(B) = 1 + 1 + ...7 terms = 7 36 36 36 ∴ Required probability = P(A/B) = P(A ∩ B) = 0 = 0. P(B) 7/36 Example 9. Consider the experiment of tossing a coin. If the coin shows head, toss it again Solution. but if it shows tail, then throw a die. Find the conditional probability of the event that ‘the die shows a number greater than 4’ given that ‘there is at least one tail’. (NCERT; CBSE 2014 C) Let S be the sample space of the experiment. ∴ S = {HH, HT, T1, T2, T3, T4, T5, T6} The outcomes of S are not equally likely. As explained in the figure, the probabilities of outcomes in S are 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 . 4 4 12 12 12 12 12 12 Let A = event that the die shows a number greater than 4 and B = event that there is at least one tail. ∴ A = {T5, T6} and B = {HT, T1, T2, T3, T4, T5, T6} 1 1 = 1 HH 2 2 4 A ∩ B = {T5, T6} H ∴ P(A ∩ B) = 1 + 1 = 1 1/2 1/4 HT 12 12 6 1 1 =112 1 1 1 1 1 1 1 6 2 T1 4 12 12 12 12 12 12 and P(B) = + + + + + + 1/12 T2 = 3 1/2 1/12T3 4 T 1/12 ∴ Required probability = P(A/B) 1/12 T4 T5 P(A ∩ B) 1/6 2 1/121/12 = P(B) = 3/4 = 9 . T6

324 Mathematics—XII Example 10. An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known: P(A fails) = 0.2, P(B fails alone) = 0.15, P(A and B fail) = 0.15. Evaluate the following probabilities: (i) P(A fails/B has failed) (ii) P(A fails alone). (NCERT) Solution. Let E = event that subsystem A fails and F = event that subsystem B fails. ∴ P(E) = 0.2, P (EF) = 0.15, P(EF) = 0.15 (∵ E F is the event that B fails alone.) Now, P(F) = P(EF or E F ) = P(EF) + P( E F ) = 0.15 + 0.15 = 0.3 (i) P(A fails/B has failed) = P(E/F) = P(EF) = 0.15 = 0.5. P(F) 0.3 (ii) P(A fails alone) = P (EF) ...(1) We have P(E) = P(EF) + P (EF) . ⇒ 0.2 = 0.15 + P (EF) ⇒ P (EF) = 0.2 – 0.15 = 0.05 ∴ (1) ⇒ P(A fails alone) = 0.05. Example 11. Mother, father and son line up at random for a family picture. Let E = event that son is on one end and F = event that father is in middle. Find P(E/F). Solution. (NCERT) Let M, F, S, respectively denote the mother, the father and the son. ∴ S = {MFS, MSF, FMS, FSM, SMF, SFM} E = {MFS, FMS, SMF, SFM} F = {MFS, SFM} E ∩ F = {MFS, SFM} ∴ P(E/F) = P(E ∩ F) = 2/6 = 1. P(F) 2/6 Example 12. Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find: (NCERT) (i) P(A ∩ B) (ii) P(A ∪ B) (iii) P(A/B) (iv) P(B/A). Solution. We have P(A) = 0.3 and P(B) = 0.4. (i) Since A and B are independent events, we have P(A ∩ B) = P(A) P(B) = 0.3 × 0.4 = 0.12. (ii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.3 + 0.4 – 0.12 = 0.58. (iii) P(A/B) = P(A ∩ B) = P(A) P(B) = P(A) = 0.3. P(B) P(B) (iv) P(B/A) = P(B ∩ A) = P(B) P(A) = P(B) = 0.4. P(A) P(A)

Probability 325 Example 13. Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6, find: (NCERT) (i) P(A and B) (ii) P(A and not B) (iii) P(A or B) (iv) P(neither A nor B). Solution. We have P(A) = 0.3 and P(B) = 0.6. (i) Since A and B are independent events, we have P(A and B) = P(A ∩ B) = P(A) P(B) = 0.3 × 0.6 = 0.18. (ii) Since A and B are independent, the events A and B are also independent. ∴ P(A and not B) = P(A ∩ B ) = P(A) P( B ) = (0.3) (1 – P(B)) = (0.3) (1 – 0.6) = 0.3 × 0.4 = 0.12. (iii) P(A or B) = P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = P(A) + P(B) – P(A) P(B) = 0.3 + 0.6 – (0.3 × 0.6) = 0.9 – 0.18 = 0.72. (iv) P(neither A nor B) = P( A ∩ B ) = P( A ) P( B ) (∵ A and B are also independent) = (1 – P(A))(1 – P(B)) = (1 – 0.3)(1 – 0.6) = 0.7 × 0.4 = 0.28. Example 14. A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even’, and B be the event ‘the number is red’. Are A and B independent? (NCERT) Solution. Here S = {1, 2, 3, 4, 5, 6} ∴ A = event that the number is even = {2, 4, 6} B = event that the number is marked in red = {1, 2, 3} ∴ A ∩ B = {2} Now, P(A) = 3 = 1 , P(B) = 3 = 1 , P(A ∩ B) = 1 6 2 6 2 6 ∴ P(A) P(B) = 1 × 1 = 1 ≠ P(A ∩ B) 2 2 4 ∴ The events A and B are not independent. Example 15. A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not. (NCERT) Solution. Here S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6} ∴ A = event that head appears on the coin = {H1, H2, H3, H4, H5, H6} B = event that 3 is on the die = {H3, T3} ∴ A ∩ B = {H3} Now, P(A) = 6 = 1 , P(B) = 2 = 1 , P(A ∩ B) = 1 12 2 12 6 12 ∴ P(A) P(B) = 1 × 1 = 1 = P(A ∩ B) 2 6 12 ∴ The events A and B are independent.

326 Mathematics—XII Example 16. A fair die is thrown two times. Let A and B be the events, ‘same number each time’, and ‘a total score is 10 or more’, respectively. Determine whether or not A and B are independent. (NCERT (EP)] Solution. Here, S = {(1, 1), (1, 2), (1, 3), ..., (6, 5), (6, 6)} A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} B = {(4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)} ∴ A ∩ B = {(5, 5), (6, 6)} P(A) = 6 = 1 , P(B) = 6 = 1 , P(A ∩ B) = 2 = 1 36 6 36 6 36 18 ∴ P(A) P(B) = 1 × 1 = 1 ≠ P(A ∩ B) 6 6 36 ∴ The events A and B are not independent. Example 17. Two dice are tossed. Find whether the following two events A and B are independent: A = {(x, y) : x + y = 11} and B = {(x, y) : x ≠ 5}, where (x, y) denotes a typical sample point. [NCERT (EP)] Solution. Here, S = {(1, 1), (1, 2), (1, 3), ..., (6, 5), (6, 6)} A = {(5, 6), (6, 5)} B = S – {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} ∴ A ∩ B = {(6, 5)} ∴ P(A) = 2 = 1 , P(B) = 36 − 6 = 5 , P(A ∩ B) = 1 36 18 36 6 36 P(A) P(B) = 1 × 5 = 5 ≠ P(A ∩ B) 18 6 108 ∴ The events A and B are not independent. Example 18. A husband and a wife appear in an interview for two vacancies for the same post. The probability of husband’s selection is 3/5 and that of wife’s selection is 1/5. Find the probability that: (i) both are selected (ii) exactly one is selected. Solution. The random experiments ‘interview of husband’ and interview of wife’ are independent. Let H = event that ‘Husband’ is selected and W = event that ‘Wife’ is selected. ∴ P(H) = 3 and P(W) = 1 5 5 Also, P(H ) = 1 – P(H) = 1– 3 = 2 and P(W ) = 1 – P(W) = 1– 1 = 4 5 5 5 5 (i) P(both are selected) = P(HW) = P(H)P(W) = 3 × 1 = 3 . 5 5 25 (ii) P(exactly one is selected) = P(HW ∪ HW )

Probability 327 = P(HW) + P(HW) (Using addition theorem) = P(H)P(W) + P(H)P(W ) = ⎛ 3 × 4 ⎞ + ⎛ 2 × 1 ⎞ = 14 . ⎝⎜ 5 5 ⎟⎠ ⎝⎜ 5 5 ⎟⎠ 25 Example 19. A speaks truth in 60% of the cases and B in 90% of the cases. In what percentage of cases, are they likely to contradict each other in stating the same fact? In the cases of contradiction do you think, the statement of B will carry more weight as he speaks truth in more number of cases than A? (CBSE 2013) Solution. The random experiments of speaking of A and B are independent. Let E = event of A speaking truth and F = event of B speaking truth ∴ P(E) = 60% = 60 = 6 and P(F) = 90% = 90 = 9 100 10 100 10 Probability of A and B contradicting each other = Probability of one speaking truth and the other not speaking truth = P(EF or EF) = P(EF) + P(EF) = P(E) P(F) + P(E) P(F) = P(E) (1 – P(F)) + (1 – P(E)) P(F) = 6 ⎛ 1 – 9 ⎞ + ⎛ 1 – 6⎞ 9 = ⎛ 6 × 1 ⎞ + ⎛ 4 × 9 ⎞ = 42 10 ⎜⎝ 10 ⎠⎟ ⎜⎝ 10 ⎠⎟ 10 ⎝⎜ 10 10 ⎟⎠ ⎜⎝ 10 10 ⎠⎟ 100 = 42% ∴ A and B are likely to contradicts each other in 42% cases. In the cases of contradiction, the statement of B will not carry more weight because contradiction occurs when either A speaks truth, B speaks false or A speaks false, B speaks truth. Example 20. A problem in Mathematics is given to 3 students whose chances of solving it are 1 , 1 , 1 . What is the probability that the problem is solved? (CBSE 2019 C) 2 3 4 Solution. The random experiments of trying the problem by the given students are independent. Let A, B, C be the event of first, second, third student solving the problem. ∴ P(A) = 1 , P(B) = 1 and P(C) = 1 2 3 4 ∴ P(problem is solved) = P(A or B or C) = I – P(AB C) = 1 – P(A) P(B) P(C ) = 1 – (1 – P(A))(1 – P(B))(1 – P(C)) = 1– ⎛ 1 − 1 ⎞ ⎛⎝⎜ 1 − 1 ⎞ ⎝⎜⎛ 1 − 1 ⎞ =1– ⎛ 1 ⎞ . ⎛ 2 ⎞ . ⎛ 3 ⎞ = 1 − 1 = 3 . ⎜⎝ 2 ⎠⎟ 3 ⎠⎟ 4 ⎠⎟ ⎜⎝ 2 ⎠⎟ ⎜⎝ 3 ⎠⎟ ⎜⎝ 4 ⎠⎟ 4 4 Example 21. Three groups of children contain 3 girls and 1 boy; 2 girls and 2 boys; 1 girl and 2 boys. One child is selected at random from each group. Show that the probability of the three selected consists of 1 girl and 2 boys is 3/8. Solution. The three random experiments of selecting one child each from group I, group II, and group III are independent.

328 Mathematics—XII 3 Girls 2 Girls 1 Girl 1 Boy 2 Boys 2 Boys Group I Group II Group III Let Bi and Gi be the events of selecting boy and girl respectively from the ith group, where i = 1, 2, 3. Required probability = P(1 girl and 2 boys) = P(G1B2B3 or B1G2B3 or B1B2G3) = P(G1B2B3) + P(B1G2B3) + P(B1B2G3) = P(G1) P(B2) P(B3) + P(B1) P(G2) P(B3) + P(B1) P(B2) P(G3) = ⎛ 3 × 2 × 2 ⎞ + ⎛ 1 × 2 × 2 ⎞ + ⎛ 1 × 2 × 1 ⎞ ⎜ + + + ⎟ ⎜ + + + ⎟ ⎜ + + + ⎟ ⎝ 3 1 2 2 1 2 ⎠ ⎝ 3 1 2 2 1 2 ⎠ ⎝ 3 1 2 2 1 2 ⎠ = ⎛ 3 × 1 × 2 ⎞ + ⎛ 1 × 1 × 2 ⎞ + ⎛ 1 × 1 × 1 ⎞ = 6 +2+ 1 = 3 . ⎝⎜ 4 2 3 ⎠⎟ ⎝⎜ 4 2 3 ⎠⎟ ⎜⎝ 4 2 3 ⎠⎟ 24 8 Example 22. A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale. (NCERT) Solution. Since the oranges are not replaced before successive draws, the three random experiments of drawing oranges are not independent. Li =et1G, 2i ,b3e. the event of drawing a good orange in the ith draw, where ∴ P(box is approved) = P(G1G2G3) 12 Good = P(G1) P(G2/G1) P(G3/G1G2) 3 Bad (By multiplication theorem) = ⎛ 12 3 ⎞ × ⎛ 12 −1 3 ⎞ × ⎛ 12 −2 3 ⎞ ⎜ 12 + ⎟ ⎜ (12 − 1) + ⎟ ⎜ (12 − 2) + ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = 12 × 11 × 10 = 44 . 15 14 13 91 Example 23. An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red? (NCERT) Solution. LinetthRe1 and Bd1r,abwe.the events of drawing red ball and black ball respectively first ∴ P(R1) = 5 = 1 and P(B1) = 5 = 1 5 Red 10 2 10 2 5 Black Let R2 be the event of drawing a red ball in the second draw.

Probability 329 ∴ Required probability = P(R2) = P(R1R2 or B1R2) = P(R1R2) + P(B1R2) = P(R1) P(R2/R1) + P(B1) P(R2/B1) = ⎛ 1 × (5 5 +2 5 ⎞ + ⎛ 1 × 5 + 5 + 2) ⎞ = 7 + 5 = 1 . ⎜ 2 + 2) + ⎟ ⎜ 2 (5 ⎟ 24 24 2 ⎝ ⎠ ⎝ ⎠ Example 24. A, B and C play a game and chances of their winning it in an attempt are 2/3, 1/2 and 1/4 respectively. A has the first chance, followed by B and then by C. The cycle is repeated till one of them wins the game. Find their respective chances of winning the game. Solution. Let Ai, Bi and Ci be the events of winning of A, B and C in their respective ith attempt, where i = 1, 2, 3, ... ∴ P(Ai) = 2 , P(Bi ) = 1 and P(Ci ) = 1 3 2 4 ∴ P(Ai ) = 1 – P(Ai) = 1– 2 = 1 , P(Bi ) = 1 – P(Bi ) = 1 – 1 = 1 3 3 2 2 and P(Ci ) = 1 – P(Ci) = 1 – 1 = 3 4 4 P(winning of A) = P(A1 or A1B1C1A2 or A1B1C1A2B2C2 A3 or ......) = P(A1) + P(A1B1C1A2 ) + P(A1B1C1A2B2C2 A3 ) + ⋅⋅⋅⋅⋅⋅ = P(A1) + P(A1) P(B1) P(C1) P(A2 ) + P(A1) P(B1) P(C1) P(A2 ) P(B2 ) P(C2 ) P(A3 ) + ⋅⋅⋅⋅⋅⋅ = 2 + ⎛ 1 × 1 × 3 × 2 ⎞ + ⎛ 1 × 1 × 3 × 1 × 1 × 3 × 2 ⎞ + ⋅⋅ ⋅ ⋅⋅ ⋅ 3 ⎝⎜ 3 2 4 3 ⎠⎟ ⎜⎝ 3 2 4 3 2 4 3 ⎟⎠ = 2 ⎡⎢1 + 1 + ⎛ 1 ⎞2 + ⎤ = 2 ⎡ 1⎤ = 2 ⎡8 ⎤ = 16 . 3 ⎣⎢ 8 ⎝⎜ 8 ⎟⎠ ⋅ ⋅ ⋅⎥ 3 ⎢⎣ 1 ⎥ 3 ⎣⎢ 7 ⎦⎥ 21 – 1/8 ⎦ ⎥⎦ P(winning of B) = P( A1 B1 or A1B1C1A2 B2 or A1B1C1A2B2C2 A3B3 or ......) = P( A1 B1) + P(A1B1C1A2B2 ) + P(A1B1C1A2B2C2 A3B3 ) + ⋅⋅...⋅ = P( A1 ) P(B1) + P( A1 ) P(B1 ) P(C1 ) P( A2 ) P(B2) + P( A1 ) P(B1 ) P(C1 ) P( A2 ) P( B2 ) P( C2 )P( A3 ) P(B3) + ⋅⋅...⋅ = ⎛ 1 × 1⎞ + ⎛ 1 × 1 × 3 × 1 × 1 ⎞ ⎝⎜ 3 2 ⎠⎟ ⎝⎜ 3 2 4 3 2 ⎟⎠ + ⎛1 × 1 × 3 × 1 × 1 × 3 × 1 × 1⎞ + ⋅⋅...⋅ ⎜⎝ 3 2 4 3 2 4 3 2 ⎠⎟

ISBN: 978-93-93738-08-0 789393 738080 T12-6743-399-COMP.CBSE QB MATHS T-II XII