Urn A contains 2 white and 8 red balls, whereas urn B contains 7 white and 2 red balls. A ball is drawn at random from urn A and placed in urn B, after which a ball from urn B was drawn and it happened to be red. What is the probability that the first ball drawn from urn A was also red? Hint: Use the Bayes formula.

I tried to solve

P(White ball from Container A) = 2/10

P(Red ball from Container A) = 8/10


So this is a pretty straight-forward application of Bayes' formula: Let $C_A$ be the color of the ball selected from urn $A$, $C_B$ the color of the ball selected from urn $B$. Then \begin{align} P[C_A = R \mid C_B = R] &= \frac{P[C_B = R, C_A = R]}{P[C_B = R]} \\ &= \frac{P[C_B = R \mid C_A = R]P[C_A=R]} {P[C_B = R \mid C_A = W]P[C_A = W] + P[C_B = R \mid C_A = R]P[C_A = R]} \\ &= \frac{(3/10)\cdot(8/10)}{(2/10)\cdot(2/10) + (3/10)\cdot(8/10)} \\ &= \frac{24}{4 + 24} \\ &= \frac{6}{7}. \end{align}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.