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?

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Solution

The correct option is **C** 8195

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∩B)=P(A)P(BA)

Now, P(A)=4C3×11C515C8

And P(BA) = Probability that the nineth examined wristwatch is defective given that there were 3 defectives in the first 8 pieces examined = 17

Hence, required probability = 4C3×11C515C8×17=8195

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∩B)=P(A)P(BA)

Now, P(A)=4C3×11C515C8

And P(BA) = Probability that the nineth examined wristwatch is defective given that there were 3 defectives in the first 8 pieces examined = 17

Hence, required probability = 4C3×11C515C8×17=8195

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