Check sibling questions

Example 5 - A die is thrown 3 times. Let A: 4 on 3rd throw B: 6 on

Example 5 - Chapter 13 Class 12 Probability - Part 2

Example 5 - Chapter 13 Class 12 Probability - Part 3 Example 5 - Chapter 13 Class 12 Probability - Part 4


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)}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), Total cases = 6 × 6 × 6 = 216 Given, A : 4 on the third throw B : 6 on the first & 5 on the second throw Thus, A ∩ B = {(6,5,4)} So, P(A ∩ B) = 1/216 A = { (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(A) = 36/216 B = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6) } P(B) = 6/216 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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.