Conditional Probability - Statement

Chapter 13 Class 12 Probability (Term 2)
Concept wise

Note:-

This question is exactly same as Example 5

### Transcript

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)} Total number = 6 × 6 × 6 = 216 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, E : 4 on the third throw F : 6 on the first & 5 on the second throw Thus, E ∩ F = {(6,5,4)} So, P(E ∩ F) = 1/216 Same as Example 5 E = { (1, 1, 4), (1, 2, 4), ……., (1, 6 ,4), (2, 1, 4), (2, 2, 4), ……., (2, 6, 4), (3, 1, 4), (3, 2, 4), ……., (3, 6, 4), (4, 1, 4), (4, 2, 4), ……., (4, 6, 4), (5, 1, 4), (5, 2, 4), ……., (5, 6, 4), (6, 1, 4), (6, 2, 4), ……., (6, 6, 4), } P(E) = 36/216 F = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6) } P(F) = 6/216 We need to find P(E|F) Now, P(E|F) = (𝑃(𝐸 ∩ 𝐹))/(𝑃(𝐹)) = (1/216)/(6/216) = 1/6 ∴ P(E|F) = 𝟏/𝟔 Same as Example 5