Question

There are $3$ bags each containing $5$ white balls and $2$ black balls and $2$ bags each containing $1$ white balls and $4$ black balls, a black ball having been drawn, find the chance that it came from the first group.

• A. $\frac{28}{43}$
• B. $\frac{15}{43}$
• C. $\frac{15}{28}$
• D. $\frac{13}{28}$

Answer 1

Answer: (B)

$\frac{15}{43}$

Total number of bags $=5$
Let A be the group in which there are $3$ bags
and B be the group in which there are $2$ bags
Now, probability of selecting bag from group A is $\frac{3}{5}$
probability of selecting bag from group B is $\frac{2}{5}$
Now, probability of selecting a black ball from first group is $\frac{3}{5}\times \frac{2}{7}=\frac{6}{35}$
Also, probability of selecting a black ball from second group is $\frac{2}{5}\times \frac{4}{5}=\frac{8}{25}$
Probability of first group , given it is black is $=\frac{\frac{6}{35}}{\frac{6}{35}+\frac{8}{25}}=\frac{15}{43}$