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Page 1 of 16, Class XI, Chapter 16 - Probability, Maths, Exercise 16.3, Question 1:, Which of the following can not be valid assignment of probabilities for outcomes of, sample space S =, {0,, 0,, 0,, 0,, O3, 0,, 0,}, Assignment, W1, W2, W3, W4, W5, W6, W7, (a), 0.1, 0.01, 0.05, 0.03, 0.01, 0.2, 0.6, (b), 1, 1, (c), 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, (d), -0.1, 0.2, 0.3, 0.4, -0.2, 0.1, 0.3, (e), 3., 4, 6., 15, 14, 14, 14, 14, 14, 14, 14, Answer, (a), W1, W2, W3, W4, W5, W6, W7, 0.1, 0.01, 0.05, 0.03, 0.01, 0.2 0.6, Here, each of the numbers p(w) is positive and less than 1., Sum of probabilities, = p(0,) + p(@.)+p(@,) + p(@.)+p(@,)+p(@,) +p(@,), = 0.1+0.01+0.05+0.03+0.01+0.2+0.6, = 1, Thus, the assignment is valid., 17

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Page 2 of 16, Class XI, Chapter 16 - Probability, Maths, (b), W1, W2, W3, W4, W5, W6, W7, 1, 1, 1, 1, 7, 7, 7, Here, each of the numbers p(wi) is positive and less than 1., Sum of probabilities, = P(@) + p(@.)+ p(@,) + p(@.) +p(@,)+p(».) +p(@,), 11.1, =-+-+ -+-+-+-+-=7x, 11.1, 7777 777, Thus, the assignment is valid., (c), W1, W2, W3, W4, W5, W6, W7, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6 0.7, Here, each of the numbers p(wi) is positive and less than 1., Sum of probabilities, =p(0,) + p(@,)+p(@,) +p(@.)+p(@,) + p(@,) +p(@,), = 0.1+0.2+0.3+0.4+0.5+0.6+0.7, = 2.8 +1, Thus, the assignment is not valid., (d), Wi, W2, W3, W4, W5, W6, W7, -0.1, 0.2, 0.3, 0.4, -0.2, 0.1, 0.3, Here, p(wi) and p(ws) are negative., Hence, the assignment is not valid., (e), W1, W2, W3, W4, W5, W6, W7, 117

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Page 3 of 16, Class XI, Chapter 16 Probability, Maths, 3, 4, 5, 15, 14, 14, 14, 14, 14, 14, 14, 15, ->1, Here, p(o,, 14, Hence, the assignment is not valid., Question 2:, A coin is tossed twice, what is the probability that at least one tail occurs?, Answer, When a coin is tossed twice, the sample space is given by, S = {HH, HT, TH, TT}, Let A be the event of the occurrence of at least one tail., Accordingly, A = {HT, TH, TT}, Number of outcomes favourable to A, Total number of possible outcomes, .:P(A)=-, n(A), n(S), 3, Question 3:, A die is thrown, find the probability of following events:, (i) A prime number will appear,, (ii) A number greater than or equal to 3 will appear,, (iii) A number less than or equal to one will appear,, (iv) A number more than 6 will appear,, (v) A number less than 6 will appear., Answer, The sample space of the given experiment is given by, S = {1, 2, 3, 4, 5, 6}, %3D, (i) Let A be the event of the occurrence of a prime number., Accordingly, A =, {2, 3, 5}

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Page 4 of 16, Class XI, Chapter 16 Probability, Maths, Number of outcomes favourable to An(A) _ 3, Total number of possible outcomes, 1, :P(A)=, %3D, n(S), 6., (ii) Let B be the event of the occurrence of a number greater than or equal to 3., Accordingly, B = {3, 4, 5, 6}, Number of outcomes favourable to B n(B) 4, Total number of possible outcomes, 2, .:P(B) =, n(S) 6, 3., (iii) Let C be the event of the occurrence of a number less than or equal to one., Accordingly, C = {1}, .: P(C)=, Number of outcomes favourable to c_ n(C) _ 1, Total number of possible outcomes, n(S) 6, (iv) Let D be the event of the occurrence of a number greater than 6., Accordingly, D = 0, Number of outcomes favourable to D n(D), :P(D) =-, =0, Total number of possible outcomes, n(S) 6, (v) Let E be the event of the occurrence of a number less than 6., Accordingly, E = {1, 2, 3, 4, 5}, %3D, ::P(E) =, Number of outcomes favourable to E n(E) 5, n(S) 6, %3D, Total number of possible outcomes, Question 4:, A card is selected from a pack of 52 cards., (a) How many points are there in the sample space?, (b) Calculate the probability that the card is an ace of spades., (c) Calculate the probability that the card is (i) an ace (ii) black card., Answer, (a) When a card is selected from a pack of 52 cards, the number of possible outcomes, is 52 i.e., the sample space contains 52 elements., Therefore, there are 52 points in the sample space., (b) Let A be the event in which the card drawn is an ace of spades., Accordingly, n(A) = 1

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Page 5 of 16, Class XI, Chapter 16 - Probability, Maths, ::P(A) =-, Number of outcomes favourable to A _n(A) _ 1, Total number of possible outcomes, %3D, n(S) 52, (c) (i)Let E be the event in which the card drawn is an ace., Since there are 4 aces in a pack of 52 cards, n(E) = 4, %3D, Number of outcomes favourable to E _n(E), Total number of possible outcomes, 4, :P(E)=-, %3D, n(S) 52, 13, (ii)Let F be the event in which the card drawn is black., Since there are 26 black cards in a pack of 52 cards, n(F) = 26, %3D, Number of outcomes favourable to F n(F) _ 26, Total number of possible outcomes n(S) 52, 1, :P(F)=, %3D, %3D, Question 5:, A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed., Find the probability that the sum of numbers that turn up is (i) 3 (ii) 12, Answer, Since the fair coin has 1 marked on one face and 6 on the other, and the die has six, faces that are numbered 1, 2, 3, 4, 5, and 6, the sample space is given by, S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6,, 6)}, Accordingly, n(S) = 12, %3D, (i) Let A be the event in which the sum of numbers that turn up is 3., Accordingly, A = {(1, 2)}, %3D, Number of outcomes favourable to A _n(A) _ 1, Total number of possible outcomes, :P(A)=, n(S) 12, (ii) Let B be the event in which the sum of numbers that turn up is 12., Accordingly, B = {(6, 6)}, Number of outcomes favourable to B n(B)_ 1, Total number of possible outcomes, .:P(B) =, n(S) 12