# 呢條數點計嫁?有冇人可以救我?

Light bulbs are produced and packed in boxes. Each box contains 15 bulbs. Under the quality control, it is found that 5% of the bulbs produced are defective.(a) If a box is selected randomly,(i) find the probability that the box contains at least one defective bulb. (ii) what are the mean and standard deviation of this probability distribution?(b) Find the probability that of 4 randomly selected boxes inspected, each contains exactly one defective bulb. (c) If a customer buys 50 boxes of bulbs, find the expected number of boxes, each of which contains exactly 2 defective bulbs.

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Probability of defective bulb = 5% = 0.05

No. of bulbs in a box = 15

This is binomial dist. with n = 15 and p = 0.05

So P(Contains at least 1 defective bulb) = 1 - P(No defective bulb)

= 1 - (15C0)(1 - 0.05)^15 = 1 - 0.95^15 = 1 - 0.46329 = 0.5367.

Mean = np = 15 x 0.05 = 0.75

Variance = nqp = 15 x 0.05 x 0.95 = 0.7125, so s.d. = sqrt 0.7125 = 0.8441.

(b)

P(Exactly 1 defective bulb in a box) = (15C1)(0.05)(1 - 0.05)^14 = 0.36576

No. of boxes = 4

So this is again a binomial dist. with n = 4 and p = 0.36576

P(All 4 boxes each containing 1 defective bulb) = (4C4)(0.36576)^4 = 0.0179.

(c)

P(Exactly 2 defective bulbs in a box) = (15C2)(0.05^2)(1 - 0.05)^13 = 0.13475

No. of boxes = 50

This is again a binomial dist. with n = 50 and p = 0.13475

So expected no. of boxes = np = 50 x 0.13475 = 6.7375