Bag A: 2 white, 1 black, 3 red.

Question

Bag A: 2 white, 1 black, 3 red. Bag B: 3 black, 2 red, $n$ white. One bag chosen at random, 2 balls drawn, found 1 red and 1 black. If $P(	ext{both from Bag A})=\dfrac{6}{11}$, find $n$.

Accepted Answer

$$P(A\mid E) = \frac{\frac{{}^3C_1{}^1C_1}{{}^6C_2}}{\frac{{}^3C_1{}^1C_1}{{}^6C_2}+\frac{{}^2C_1{}^3C_1}{{}^{n+5}C_2}} = \frac{6}{11}$$

Solving gives $n=4$.

Solutions · 1