Question -
Answer -
Let us assume that thenumber of bulbs that will fuse after 150 days of use in an experiment of 5trials be x.
As we can see that thetrial is made with replacement, thus, the trials will be Bernoulli trials.
It is alreadymentioned in the question that, p = 0.05
Thus, q = 1 – p = 1 –0.05 = 0.95
Here, we can clearlyobserve that x has a binomial representation with n = 5 and p = 0.05
Thus, P(X = x) = nCx qn-x px,where x = 0, 1, 2… n
= 5Cx (0.95)5-x(0.05)x
(i) Probability of nosuch 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 ofnot 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 ofmore 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 atleast 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