The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs

(i) none

(ii) not more than one

(iii) more than one

(iv) at least one will fuse after 150 days of use.

Let us assume that the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials be x.

As we can see that the trial is made with replacement, thus, the trials will be Bernoulli trials.

It is already mentioned in the question that, p = 0.05

Thus, q = 1 – p = 1 – 0.05 = 0.95

Here, we can clearly observe that x has a binomial representation with n = 5 and p = 0.05

Thus, P(X = x) = ^{n}C_{x}q^{n-x}p^{x} , where x = 0, 1, 2, …n

= ^{5}C_{x}(0.95)^{5-x}(0.05)^{x}

(i) Probability of no such bulb in a random drawing of 5 bulbs = P(X = 0)

= ^{5}C_{0}(0.95)^{5-0}(0.05)^{0}

= 1× 0.95^{5}

= (0.95)^{5}

(ii) Probability of not more than one such bulb in a random drawing of 5 bulbs = P(X≤ 1)

= P(X = 0) + P(X = 1)

= ^{5}C_{0}(0.95)^{5-0}(0.05)^{0}+ ^{5}C_{1}(0.95)^{5-1}(0.05)^{1}

= 1× 0.95^{5} + 5 × (0.95)^{4} × 0.05

= (0.95)^{4} (0.95 +0.25)

= (0.95)^{4} × 1.2

(iii) Probability of more than one such bulb in a random drawing of 5 bulbs = P(X>1)

= 1 – P(X ≤ 1)

= 1 – [(0.95)^{4} × 1.2]

(iv) Probability of at least one such bulb in a random drawing of 5 bulbs = P(X ≥ 1)

= 1 – P(X < 1)

= 1 – P(X = 0)

= 1 –(0.95)^{5}

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