Question

If the probability that a florescent light has a useful life of at least 800 hours is 0.9,
find the probabilities that among 20 such lights
(a). exactly 18 will have a useful life of at least 800 hours;
(b). exactly 15 will have a useful life of at least 800 hours;
(c). at least 2 will not have a useful life of at least 800 hours;

Answer 1

p=0.9

n=20

The binomial distribution

$P(X=x) = \binom{n}{x}(p)^x(1-p)^{n-x}$

(a) exactly 18 will have a useful life of at least 800 hours

$P(X=18) = \binom{20}{18}(0.9)^{18}(1-0.9)^{20-18} \\ = \frac{20!}{18!(20-18)!} \times (0.9)^{18} \times (0.1)^2 \\ = 0.285179$

(b) exactly 15 will have a useful life of at least 800 hours

$P(X=15) = \binom{20}{15}(0.9)^{15}(1-0.9)^{20-15} \\ = \frac{20!}{15!(20-15)!} \times (0.9)^{15} \times (0.1)^5 \\ = 0.031921$

(c) at least 2 will not have a useful life of at least 800 hours

$p=1-0.9=0.1 \\ P(X≥2) = 1 -P(X<2) \\ = 1-[P(X=0)+P(X=1)] \\ = 1 -[\binom{20}{0}(0.1)^0(0.9)^{20-0} + \binom{20}{1} (0.1)^1 (0.9)^{20-1}] \\ = 1-[0.121576 + 0.270170] \\ = 1 -0.391746 \\ = 0.608254$