Three defective bulbs are mixed with 8 good ones. If three bulbs are drawn one by one with replacement, the probabilities of getting exactly 1 defective, more than 2 defective, no defective and more than 1 defective respectively are:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{576}{1331}, \frac{27}{1331}, \frac{512}{1331}$ and $\frac{243}{1331}$
Total bulbs $=11$, defective $=3$, good $=8$
Probability of defective $p=\frac{3}{11}$, good $q=\frac{8}{11}$
Three draws with replacement ⇒ binomial with $n=3$
Probability of exactly $1$ defective
$P(X=1)=\frac{3!}{1!2!}p q^2 =3\cdot\frac{3}{11}\cdot\left(\frac{8}{11}\right)^2 =\frac{576}{1331}$
Probability of more than $2$ defective means $X=3$
$P(X=3)=p^3=\left(\frac{3}{11}\right)^3=\frac{27}{1331}$
Probability of no defective
$P(X=0)=q^3=\left(\frac{8}{11}\right)^3=\frac{512}{1331}$
Probability of more than $1$ defective means $X=2$ or $3$
$P(X\ge2)=P(X=2)+P(X=3)$
$P(X=2)=\frac{3!}{2!1!}p^2 q =3\cdot\left(\frac{3}{11}\right)^2\cdot\frac{8}{11} =\frac{216}{1331}$
$P(X\ge2)=\frac{216}{1331}+\frac{27}{1331}=\frac{243}{1331}$
The required probabilities are
Exactly $1$ defective $=\frac{576}{1331}$,
More than $2$ defective $=\frac{27}{1331}$,
No defective $=\frac{512}{1331}$,
More than $1$ defective $=\frac{243}{1331}$.