Answer:
Given
that p = 0.05
q = 1 ?
0.05 = 0.95
Let
x be the number of fused bulbs, then x has the binomial distribution with n =
5, p = 0.5, q = 0.95
(i)
P(none of fused bulb) = P (x = 0)
=
5C0 (0.95)5-0 (0.05)0 = (0.95)5
(ii)
P (not more than one fused bulb)
=
5C0 (0.95)5-0 (0.05)0 + 5C1
(0.05)5?1 (0.05)1
=
(0.95)4 + 5 (0.95)4 (0.05)
=
(0.95)4 (0.95 + 0.25)
=
1.2 × (0.95)4
(iii)
P (more than one fused bulbs)
=
P(x > 1) = 1 [P (x = 0) + P(x = 1)]
=
1 ? (1.2) (0.95)4 [Using (ii) part]
(iv)
P (atleast one fused blub) = P (x 1)
=
1 ? P (x < 1) =1 ? P (x = 0)
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