Question

A consignment of 15 wristwatches contains 4 defectives. The wristwatches are selected at random, one by one and examined. The ones examined are not put back. What is the probability that ninth one examined is the last defective?

• A. $\frac{11}{195}$
• B. $\frac{17}{195}$
• C. $\frac{8}{195}$
• D. $\frac{16}{195}$

Answer 1

The correct option is C $\frac{8}{195}$

Let A be the event of getting exactly 3 defectives in the examination of 8 wristwatches.
And B be the event of getting ninth wristwatch defective.
Then
Required probability = $(P\cap B)=P\left(A\right)P\left(\frac{B}{A}\right)$
Now, $P\left(A\right)=\frac{{}^4{C}_3{\times }^{11}{C}_5}{{}^{15}{C}_8}$
And $P\left(\frac{B}{A}\right)$ = Probability that the nineth examined wristwatch is defective given that there were 3 defectives in the first 8 pieces examined = $\frac{1}{7}$
Hence, required probability = $\frac{{}^4{C}_3{\times }^{11}{C}_5}{{}^{15}{C}_8}\times \frac{1}{7}=\frac{8}{195}$