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
Best Answer
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}