Ex 13.1, 8 - Chapter 13 Class 12 Probability

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Sample Space When die is thrown 3 timesA die is thrown 3 times S = {(1, 1, 1), (1, 1, 2), ......, (1, 1, 6), (1, 2, 1), (1, 2, 2), ......, (1, 2, 6), (1, 3, 1), (1, 3, 2), ......, (1, 3, 6), (1, 4, 1), (1, 4, 2), ......, (1, 4, 6), (1, 5, 1), (1, 5, 2), ......, (1, 5, 6), (1, 6, 1), (1, 6, 2), ......, (1, 6, 6), (2, 1, 1), (2, 1, 2), ......, (2, 1, 6), (2, 2, 1), (2, 2, 2), ......, (2, 2, 6), (2, 3, 1), (2, 3, 2), ......, (2, 3, 6), (2, 4, 1), (2, 4, 2), ......, (2, 4, 6), (2, 5, 1), (2, 5, 2), ......, (2, 5, 6), (2, 6, 1), (2, 6, 2), ......, (2, 6, 6), (3, 1, 1), ......, (3, 1, 6), (3, 2, 1),......, (3, 2, 6), (3, 3, 1),......, (3, 3, 6), (3, 4, 1),......, (3, 4, 6), (3, 5, 1),......, (3, 5, 6), (3, 6, 1),......, (3, 6, 6), (4, 1, 1), ……………..(4, 6, 6), (5, 1, 1), ……………..(5, 6, 6), (6, 1, 1), ……………..(6, 6, 6)} Ex 13.1, 8 A die is thrown three times, E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses Determine P(E|F) A die is thrown 3 times S = {(1, 1, 1) ,.........., (1, 6, 6), (2, 1, 1), .........., (2, 6, 6), (3, 1, 1), .........., (3, 6, 6), (4, 1, 1), ……………..(4, 6, 6), (5, 1, 1), ……………..(5, 6, 6), (6, 1, 1), ……………..(6, 6, 6), Total cases = 6 × 6 × 6 = 216 Given, A: 4 on the third throw B: 6 on the first & 5 on the second throw We need to find the probability of A, given that B has already occurred i.e. P(A|B) Finding P(B) B = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6) } P(B) = 6/216 Finding P(A ∩ B) A ∩ B = {(6, 5, 4)} P(A ∩ B) = 1/216 Now, P(A|B) = (𝑃(𝐴 ∩ 𝐵))/(𝑃(𝐵)) = (1/216)/(6/216) = 𝟏/𝟔