A Consignment of 12 mobile phones contains 4

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