Q. 5

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.

Answer :

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) = nCxqn-xpx , where x = 0, 1, 2, …n


= 5Cx(0.95)5-x(0.05)x


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


= 5C0(0.95)5-0(0.05)0


= 1× 0.955


= (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)


= 5C0(0.95)5-0(0.05)0+ 5C1(0.95)5-1(0.05)1


= 1× 0.955 + 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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