Question
A Consignment of 12 mobile phones contains 4
defectives. The mobile phones are selected at random one by one and examined. The ones examined are not put back. What is the probability that seventh one examined is the last defective?Solution
let A be the event of getting exactly 3 defectives in the examination of six mobile phones And B be the event of getting seventh mobile phone is defective. Then, Required Probability = P(A∩B) = P(A) × P(B/A) Now, P(A) = 4C3 × 8C3 / 12C6 = (4!/3! × 8!/(5! ×3!))/(12!/(6! ×6!)) = (4 ×56)/924 = 8/33 And P(B/A) = Probability that the seventh examined mobile phones is defective given that there were three defectives in the first six pieces examined = 1/6 Hence, the Required Probability = 8/33 ×1/6 = 4/99
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