Question

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

The correct option is D $\frac{3}{7}$

The probability of choosing the first box is $\frac{3}{6}=\frac{1}{2}$

The probability of choosing the first box is $\frac{3}{6}=\frac{1}{2}$

probability of choosing the black ball is (The probability of choosing the first box $\times$ probability of choosing the black) + (The probability of choosing the first box $\times$ probability of choosing the black)

$=(\frac{1}{2}\times \frac{3}{7})+(\frac{1}{2}\times \frac{4}{7})$

= $\frac{1}{2}$

The required probability = $\frac{\left(Pballchoosenfromfirstbox\right)\times \left(choosenballisblack\right)}{\left(Pchoosenballisblack\right)}$

= $\frac{(\frac{1}{2}\times \frac{3}{7})}{\frac{1}{2}}=\frac{3}{7}$.