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:

$\frac{576}{1331}, \frac{27}{1331}, \frac{512}{1331}$ and $\frac{243}{1331}$

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 $=\frac{243}{1331}$.