From a lot of 15 bulbs which include 5 defective bulbs, a sample of 4…

The correct answer is Option (1) → 

Probability Distribution:

$X$	0	1	2	3	4
$P(X)$	$\frac{16}{81}$	$\frac{32}{81}$	$\frac{24}{81}$	$\frac{8}{81}$	$\frac{1}{81}$

Mean: $\frac{4}{3}$

Total number of bulbs in the lot = 15,

number of defective bulbs = 5.

Let the probability of a defective bulb drawn be p, then $p =\frac{5}{15}=\frac{1}{3}$

∴ the probability of a good bulb = $q = 1-p=1-\frac{1}{3}=\frac{2}{3}$.

As 4 bulbs are drawn one by one with replacement, the events are independent therefore it is a problem of binomial distribution with $p =\frac{1}{3},q=\frac{2}{3}$ and $n = 4$.

Let random variable X denote the number of defective bulbs drawn, then X can take values 0, 1, 2, 3, 4.

$P(X = 0) = {^4C}_0q^42 = 1×(\frac{2}{3})^4=\frac{16}{81},$

$P(X = 1) = {^4C}_1 pq^3 = 4×\frac{1}{3}×(\frac{2}{3})^3= \frac{32}{81},$

$P(X = 2)={^4C}_2 p^2q^2 = 6×(\frac{1}{3})^2×(\frac{2}{3})^2= \frac{24}{81},$

$P(X = 3) = {^4C}_3 p^3q = 4×(\frac{1}{3})^3×\frac{2}{3} =\frac{8}{81}$ and

$P(X = 4) = {^4C}_4p^4 = 1×(\frac{1}{3})^4=\frac{1}{81}$

∴ The probability distribution of X is $\begin{pmatrix}0&1&2&3&4\\\frac{16}{81}&\frac{32}{81}&\frac{24}{81}&\frac{8}{81}&\frac{1}{81}\end{pmatrix}$.

Mean = $Σp_ix_i = \frac{1}{81}(16 × 0+32 × 1 + 24 × 2 + 8 × 3+1 × 4)$

$=\frac{1}{81}(0 +32 +48 +24+4)= \frac{108}{81}=\frac{4}{3}$