From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by…

Question

From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is

Accepted Answer

total number of defective bulbs $n(D)=10$

total number of non defective bulbs $n(\vec{D})=90$

total bulbs $=n(D)+n(\vec{D})=100$

probability of picking a defective bulbs: $p=\frac{10}{100}=\frac{1}{10}$

probability of picking a non-defective bulbs $q=\frac{90}{100}=\frac{9}{10}$

number of trials $n=8$ bulbs are selected

let $X$ be the event of getting atleast 7 defective bulbs

$\begin{gathered} P(X)=P(X=7)+P(X=8) \ ={}^{8}{C}_{7}{p}^{7}\cdot {q}^{1}+{}^{8}{C}_{8}{p}^{8}{q}^{0} \ =8\cdot {(\frac{1}{10})}^{7}\cdot \frac{9}{10}+1\cdot {(\frac{1}{10})}^{8} \end{gathered}$

$P(X)={(\frac{1}{10})}^{8}[72+1]=\frac{73}{{10}^{8}}$

Solutions · 1