The probability that a bulb produced by a factory will fuse after 6 months of use is 0.05. find the probability that out of 5 such bulbs
(i) none will fuse after 6 months of use
(ii) at least one will fuse after 6 months of use
(iii) not more than one will fuse after 6 months of use
(i) The probability that the bulb will fuse = 0.05 = p
The probability that the bulb will not fuse = 1-0.05 = 0.95 = q
Using Bernoulli’s we have,
P(Success=x) = nCx.px.q(n-x)
x=0, 1, 2, ………n and q = (1-p), n =5
Probability that none will fuse =
5C0.(0.05)0(0.95)5
⇒ (0.95)5
(ii) The probability that the bulb will fuse = 0.05 = p
The probability that the bulb will not fuse = 1-0.05 = 0.95 = q
Using Bernoulli’s we have,
P(Success=x) = nCx.px.q(n-x)
x=0, 1, 2, ………n and q = (1-p), n =5
Probability that at least one will fuse = P(1) + P(2) + P(3) + P(4) + P(5)
5C1.(0.05)1(0.95)4 + 5C2.(0.05)2(0.95)3 + 5C3.(0.05)3(0.95)2 + 5C4.(0.05)4(0.95)1 + 5C5.(0.05)5(0.95)0
⇒ (1-(0.95)5)
(iii) The probability that the bulb will fuse = 0.05 = p
The probability that the bulb will not fuse = 1-0.05 = 0.95 = q
Using Bernoulli’s we have,
P(Success=x) = nCx.px.q(n-x)
x=0, 1, 2, ………n and q = (1-p), n =5
Probability that not more than one will fuse = P(0) + P(1)
5C0.(0.05)0(0.95)5 + 5C1.(0.05)1(0.95)4
⇒ (1.20).(0.95)5