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
11195
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
17195
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
8195
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
16195
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is C 8195Let 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

Suggest Corrections
0
Related Videos
QUANTITATIVE APTITUDE
Watch in App
Explore more