Question

Four boxes $A. B. C$ and $D$ contain $5000,3000,2000$ and 1000 fuses respectively. The percentages of defective fuses in these boxes are $3 \%, 2 \%, 1 \%$ and $0.5 \%$ respectively. If a fuse selected at random from one of the boxes is found to be defective, then the probability that it has come from box $D$ is

• A. $\frac{1}{13}$
• B. $\frac{4}{65}$
• C. $\frac{1}{65}$
• D. $\frac{2}{13}$

Answer 1

Answer: (A)

Consider the event
$A=$ Box $A$ is selected
$B=$ Box $B$ is selected
$C=$ Box $C$ is selected
$D=$ Box $D$ is selected
$E=$ Defective fuses
$p(A)=\frac{5000}{11000}=\frac{5}{11}$
$p(B)=\frac{3}{11}, p(C)=\frac{2}{11}$
$p(D)=\frac{1}{11}, p\left(\frac{E}{A}\right)=\frac{3}{100}$
$p\left(\frac{E}{B}\right)=\frac{2}{100}, p\left(\frac{E}{C}\right)$
$=\frac{1}{100}, p\left(\frac{E}{D}\right)=\frac{1}{200}$
$\therefore $ Required probability
$=p\left(\frac{D}{E}\right)=\frac{p(D) \times p\left(\frac{E}{D}\right)}{p(A) \times p\left(\frac{E}{A}\right)+p(B) \times p\left(\frac{E}{B}\right)}$
$+p(C) \times p\left(\frac{E}{C}\right)+p(D) \times p\left(\frac{E}{D}\right)$
$\frac{1}{11} \times \frac{1}{200}$
$=\frac{5}{11} \times \frac{2}{100}+\frac{3}{11} \times \frac{2}{100}+\frac{2}{11} \times \frac{1}{100}+\frac{1}{11} \times \frac{1}{200}=\frac{1}{47}$