Question

There are $4$ white and $3$ black balls in a box. In another box there are $3$ white and $4$ black balls. An unbiased dice is rolled. If it shows a number less than or equal to $3$, then a ball is drawn from the first box, but if it shows a number more than $3$, then a ball is drawn from the second box. If the ball drawn is black, then the probability that the ball was drawn from the first box is

• A. $\frac{1}{2}$
• B. $\frac{6}{7}$
• C. $\frac{4}{7}$
• D. $\frac{3}{7}$

Answer 1

Answer: (D)

$\frac{3}{7}$

Probability of choosing the first box $=\frac{3}{6}=\frac{1}{2}$
Probability of choosing the second box $=\frac{3}{6}=\frac{1}{2}$
Probability of choosing a black ball
$=$(Probability of choosing the first bag * Probability of choosing black)
$+$ (Probability of choosing the second bag * Probability of choosing black)
$=(\frac{1}{2}\times \frac{3}{7})+(\frac{1}{2}\times \frac{4}{7})$
$=\frac{1}{2}$
Now,
$P\left(\frac{ballchosenfromfirstbox}{chosenballisblack}\right)$
$=\frac{P\left(\right(ballchosenfromfirstbox)\bigcap (chosenballisblack\left)\right)}{P\left(chosenballisblack\right)}$
Thus, required probability = $\frac{(\frac{1}{2}\times \frac{3}{7})}{\frac{1}{2}}=\frac{3}{7}$