✔ 最佳答案
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