A box of bulbs is inspected by examining three randomly

Solution

Let A,B, and C be the respective events that the first, second, and third drawn orange is non-defected. Therefore, probability that first drawn orange is non-defected, P(A)=12/15 The bulbs are not replaced. Therefore, probability of getting second orange non-defected, P(B) = 11/14 Similarly, probability of getting third orange non-defected, P(C)= 10/13 The box is approved for sale if all the three bulbs are non-defected. Thus, probability of getting all the bulbs non-defected = 12/15 × 11/14 × 10/13 = 44/91 = 0.48