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,
will fuse after 150 days of use.
Let P be probability of success and q be the probability of failure
⇒ P = 0.05, q = 0.95
n = 5
(i)n(x=0)=nC1p0qn−0=nC0q5=(0.95)5(ii)p(x≤1)=p(x=0)+p(x=1)=5C0p0q5+5C1p1q4=(0.95)5+5(0.05)(0.95)4