Question

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.

Answer 1

Let P be probability of success and q be the probability of failure

$\Rightarrow$ P = 0.05, q = 0.95

n = 5

$\left(i\right)n(x=0)=n_{C_1}{p}^0{q}^{n-0}=n_{C_0}q5=\left(0.95{)}^5\right(ii\left)p\right(x\le 1)=p(x=0)+p(x=1)=5_{C_0}{p}^0{q}^5+5_{C_1}{p}^1{q}^4=(0.95{)}^5+5\left(0.05\right)(0.95{)}^4$