Answer to Question #323689 in Statistics and Probability for Klum

Question

Two boxes, 1 and 2 are given. Suppose box 1 contains 6 black and 7 white balls. Box 2 contains 9 black and 8 white balls. Three balls are selected at random from box 1 and put into box 2. Then two balls are selected at random from box 2.

1) what is the probability that these two balls have the same color

2) what is the probability that these two balls have the different colors

Answer 1

$1:\\H_1:www\,\,from\,\,box\,\,1\\H_2:wwb\,\,from\,\,box\,\,1\\H_3:wbb\,\,from\,\,box\,\,1\\H_4:bbb\,\,from\,\,box\,\,1\\A: same\,\,colors\\P\left( H_1 \right) =\frac{C_{7}^{3}}{C_{13}^{3}}=0.122378\\P\left( H_2 \right) =\frac{C_{7}^{2}C_{6}^{1}}{C_{13}^{3}}=0.440559\\P\left( H_3 \right) =\frac{C_{7}^{1}C_{6}^{2}}{C_{13}^{3}}=0.367133\\P\left( H_4 \right) =\frac{C_{6}^{3}}{C_{13}^{3}}=0.0699301\\P\left( A|H_1 \right) =\left[ 9b,11w \right] =\frac{C_{11}^{2}+C_{9}^{2}}{C_{20}^{2}}=0.478947\\P\left( A|H_2 \right) =\left[ 10b,10w \right] =\frac{C_{10}^{2}+C_{10}^{2}}{C_{20}^{2}}=0.473684\\P\left( A|H_3 \right) =\left[ 11b,9w \right] =\frac{C_{9}^{2}+C_{11}^{2}}{C_{20}^{2}}=0.478947\\P\left( A|H_4 \right) =\left[ 12b,8w \right] =\frac{C_{8}^{2}+C_{12}^{2}}{C_{20}^{2}}=0.494737\\P\left( A \right) =P\left( A|H_1 \right) P\left( H_1 \right) +P\left( A|H_2 \right) P\left( H_2 \right) +P\left( A|H_3 \right) P\left( H_3 \right) +P\left( A|H_4 \right) P\left( H_4 \right) =\\=0.478947\cdot 0.122378+0.473684\cdot 0.440559+0.478947\cdot 0.367133+0.494737\cdot 0.0699301=\\=0.477733\\2: P\left( different \right) =1-P\left( same \right) =1-0.477733=0.522267$