Answer to Question #331779 in Statistics and Probability for ash

Question

 Jar A contains one white and two black balls, jar B contains five white and two black balls, and jar C contains four white and two black balls. One ball is selected from each jar. What is the probability that the ball chosen from jar B will be white, given that exactly two black balls are selected?

Show you work.

Answer 1

Let's denote W a white ball, B a black ball.

Jar A contains 3 balls: 1 W, 2 B;

Jar B contains 7 balls: 5 W, 2 B;

Jar C contains 6 balls: 4 W, 2 B.

Let X be an event that exactly two black balls are selected after selecting a ball from each jar,

X=H₁ $\cup$ H₂ $\cup$ H₃, where

H₁ = WBB (a white ball is selected from the jar A, black from B and C),

H₂ = BWB (a white ball is selected from the jar B, black from A and C),

H₃ = BBW (a white ball is selected from the jar C, black from A and B).

Then we are to find the value of P(H₂|X).

Using Bayes’ theorem formula we get:

$P(H_2|X)=\\ =\cfrac{P(X|H_2)P(H_2)}{P(X|H_1)P(H_1)+P(X|H_2)P(H_2)+P(X|H_3)P(H_3)};$

$P(H_1)=\cfrac{1}{3}\cdot\cfrac{2}{7}\cdot\cfrac{2}{6}=\cfrac{2}{63};\\ P(H_2)=\cfrac{2}{3}\cdot\cfrac{5}{7}\cdot\cfrac{2}{6}=\cfrac{10}{63};\\ P(H_3)=\cfrac{2}{3}\cdot\cfrac{2}{7}\cdot\cfrac{4}{6}=\cfrac{8}{63};\\ P(X|H_1)=P(X|H_2)=P(X|H_3)=1;\\ P(H_2|X)=\\ =\cfrac{1\cdot\cfrac{10}{63}}{1\cdot\cfrac{2}{63}+1\cdot\cfrac{10}{63}+1\cdot\cfrac{8}{63}}=\cfrac{1}{2}.$