Question

Bag $I$ contains $3$ red and $4$ black balls, while another bag $II$ contains $5$ red and $6$ black balls. One ball is drawn at random from one of the bags and it is found to be black. The probability that it was drawn from bag $II$ is

• A. $\frac{7}{43}$
• B. $\frac{13}{43}$
• C. $\frac{21}{43}$
• D. None of these

Answer 1

Answer: (C)

$\frac{21}{43}$

Answer 2

The probability of black ball drawn from bag 1st
$=\frac{4}{3+4}=\frac{4}{7}$
The probability of black ball drawn from bag Ilnd
$=\frac{6}{5+6}=\frac{6}{11}$
Total probability $=\frac{4}{7}+\frac{6}{11} =\left(\frac{86}{77}\right)$
So, required probability
$=\frac{\left(6 /11\right)}{\left(86/ 77\right)}$
$=\frac{6 \times77}{86\times11}$
$=\frac{21}{43}$