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