Question

There are 3 bags which are known to contain 2 white and 3 black balls; 4 white and 1 black balls and 3 white and 7 black balls respectively. A ball is drawn at random from one of the bags and found to be a black ball. Then the probability that it was drawn from the bag containing the most black balls is

• A. 7/15
• B. 5/19
• C. 3/4
• D. None of these

Answer 1

The correct option is A

7/15

consider the following events:

$A\to Balldrawnisblack$; $E_1\to BagIischosen$;

$E_2\to BagIIischosen$ and $E_3\to BagIIIischosen$

$P\left(E_1\right)=\left(E_2\right)=P\left(E_3\right)=\frac{1}{3},P\left(\frac{A}{E_1}\right)=\frac{3}{5}$

Then

$P\left(\frac{A}{E_2}\right)=\frac{1}{5},P\left(\frac{A}{E_3}\right)=\frac{7}{10}$

Required probability is $=P\left(\frac{E_3}{A}\right)$

$=\frac{P\left(E_3\right)P\left(\frac{A}{E_3}\right)}{P\left(E_1\right)P\left(\frac{A}{E_1}\right)+P\left(E_2\right)P\left(\frac{A}{E_2}\right)+P\left(E_3\right)P\left(\frac{A}{E_3}\right)}=\frac{7}{15}$