Question

A bag contains 4 white and 5 black balls. A second bag contains 3 (identical) white and 6 (identical) black balls. One bag is chosen at random and a ball is drawn. Its colour is noted and the ball replaced. This is repeated four times. It was found that of these four, one was white and 3 were black. The $n$ the probability that the first bag was chosen is

• A. $\frac{162}{287}$
• B. $\frac{125}{287}$
• C. $\frac{250}{280}$
• D. $\frac{125}{574}$

Answer 1

The correct option is B $\frac{125}{287}$

If $A$ is the event of choosing the first bag and $B$ the event of choosing the second bag. $P\left(A\right)=P\left(B\right)=12$

Let $C$ be the event of one white and $3$ black being drawn .

Using Baye's theorem

$P\left(A|C\right)=\frac{P\left(C|A\right)\ast P\left(A\right)}{P\left(C|A\right)\ast P\left(A\right)+P\left(C|B\right)\ast P\left(B\right)}=\frac{1/2\times (4/9{)}^2\times (5/9{)}^3}{1/2\times (4/9{)}^2\times (5/9{)}^3+1/2\times (1/3{)}^2\times (2/3{)}^3}=125/287$