# Probability Box 1 contains 2 red balls and 1 blue ball. Box 2 contains 3 blue balls and 1 red ball. A coin is tossed.

Box 1 contains 2 red balls and 1 blue ball. Box 2 contains 3 blue balls and 1 red ball. A coin is tossed. If it falls heads up, box 1 is selected and a ball is drawn. If it falls tails up, box 2 is selected and a ball is drawn. If the ball is red, what is the probability it came from box 1? (The given answer is 8/11 but i have no idea how they got it)

• Consider using Bayes' Theorem. Feb 1, 2016 at 0:54

Number the balls, so that box $1$ contains $\{R_{1,1},R_{1,2},B_1\}$ and box $2$ contains $\{B_{2,1},B_{2,2},B_{2,3},R_2\}$
The probability of drawing $R_{1,1}$ is $\frac 12 \times \frac 13=\frac 16$, this is also the probability of drawing $R_{1,2}$.
The probability of drawing $R_2$ is $\frac 12 \times \frac 14= \frac 18$.
Thus the probability of drawing a red ball is $\frac 26 +\frac 18=\frac {11}{24}$. The probability of drawing a red ball from box $1$ is $\frac 26=\frac 13$.
Thus the answer is $$\frac {\frac 13}{\frac {11}{24}}=\frac 8{11}$$