A box contains 16 transistors, 5 of which are defective. if 5 are selected at random, find the probability thata. all are defective. and. none are defective.
Accepted Solution
A:
Part (a): All are defective Only one way of selecting the 5 defective transistors:
Number of ways of selections available = 6C5 = 16!/[5!*(15-5)!] = 4368
Probability they are all defective = Number of ways of selecting 5 defectives/Total number of ways possible = 1/4368 ≈ 0.000229
Part (b): None are defective Total number of non defectives = 16 -5 = 11 Number of ways of selecting 5 non defective = 11C5 = 462 ways Total number of ways possible = 16C5 = 4368 Probability of selecting 5 non defectives = 462/4368 = 11/104 ≈ 0.1058