Example 5 - Class 12

Transcript

Example 5 (Introduction) A 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), (3,1,2), .........., (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), Example 5 A die is thrown three times. Events A and B are defined as below: A : 4 on the third throw B : 6 on the first and 5 on the second throw Find the probability of A given that B has already occurred. 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), Given, A : 4 on the third throw B : 6 on the first & 5 on the second throw A ∩ B = {(6,5,4)} So, P(A ∩ B) = 1﷮216﷯ We need to find the probability of A, given that B has already occurred i.e. P(A|B) Now, P(A|B) = 𝑃(𝐴 ∩ 𝐵)﷮𝑃(𝐵)﷯ = 1﷮216﷯﷮ 6﷮216﷯﷯ = 1﷮6﷯ ∴ P(A|B) = 𝟏﷮𝟔﷯