Question

There are $2$ bags one of which contains $3$ black and $4$ white balls, while the other contains $4$ black and $3$ white balls. A die is cast, if face $1$ or $3$ turns up a ball in taken from the $1$st bag and if any other face turns up a ball is taken from the second bag. The probability of choosing a black ball is

• A. $\frac{5}{21}$
• B. $\frac{10}{21}$
• C. $\frac{11}{21}$
• D. $\frac{6}{21}$

Answer 1

Answer: (C)

$\frac{11}{21}$

Bag $1$: $3B+4W=7$ balls
Bag $2$: $4W+3B=7$ balls
Selection of a Bag $1$ is$=\frac{2}{6}$ (when $1$ or $3$ face turns up)
Selection of a Bag $2$ is$=\frac{4}{6}$ (when $2,4,5$ or $6$ face turns up)
Probability that the one drawn ball is white$=(\frac{2}{6}\times \frac{3}{7})+(\frac{4}{6}\times \frac{4}{7})=\frac{11}{21}$