Bag I: 3 red, 4 black, 3 white.

Question

Bag I: 3 red, 4 black, 3 white. Bag II: 2 red, 5 black, 2 white. One ball transferred from I to II, then a ball drawn from II is black. Find the probability the transferred ball was red.

Accepted Answer

$$P(\text{red transferred}\mid\text{black drawn}) = \frac{\frac{3}{10}\times\frac{5}{10}}{\frac{3}{10}\times\frac{5}{10}+\frac{4}{10}\times\frac{6}{10}+\frac{3}{10}\times\frac{5}{10}} = \frac{15}{15+24+15}=\frac{15}{54}=\frac{5}{18}$$

Solutions · 1