Question

The probability that a bulb produced by a factory will fuse after $100$ days of use is $0.05$. Find the probability that out of $5$ such bulbs,
i) none.

Answer 1

P (that a bulb will fuse after 150 days) $=p=\mathrm{0.05.}$

P (that a bulb won't fuse after 150 days) $=q=1-p=0.95$

Here $n=5,p=0.05,q=0.95$

(i)none

$P(X=0){=}^5{C}_0\times (0.05{)}^0\times (0.95{)}^5={0.95}^5$

(ii) not more than one

$P(X\le 1)=P(X=0)+P(X=1)$

$P(X\le 1)={0.95}^5{+}^5{C}_1\times (0.05{)}^1\times (0.95\left)\right(5-1)={0.95}^5+{0.95}^4(5\times 0.05)$

$P(X\le 1)={0.95}^4(0.95+0.25)=1.2\times {0.95}^4$

(iii) more than one

$P(X>1)=1-P(X\le 1)=1-1.2\times {0.95}^4$

(iv)at least one

$P(X\ge 1)=1-P(X=0)=1-{0.95}^5$