Four bad eggs are mixed with 10 good ones. If three eggs are drawn one by one with replacement, then find the probability distribution of the number of bad eggs drawn.

The correct answer is Option (3) → 

Since the eggs are drawn one by one with replacement, the events are independent, therefore, it is a problem of binomial distribution.

Total number of eggs = 4 + 10 = 14, out of which 4 are bad.

If $p$ = probability of drawing a bad egg, then $p =\frac{4}{14}=\frac{2}{7}$, so $q=1-\frac{2}{7}=\frac{5}{7}$.

Thus, we have a binomial distribution with $p = \frac{2}{7},q = \frac{5}{7}$ and $n = 3$.

If X denotes the number of bad eggs obtained, then X can take values 0, 1, 2, 3.

$P(0) = {^3C}_0 q^3 = (\frac{5}{7})^3=\frac{125}{343}$

$P(1)={^3C}_1pq^2 = 3.\frac{2}{7}.(\frac{5}{7})^3=\frac{150}{343}$,

$P(2) = {^3C}_2 p^2q = 3. (\frac{2}{7})^2.\frac{5}{7}=\frac{60}{343}$ and

$P(3) = {^3C}_3 p^3 = (\frac{2}{7})^3=\frac{8}{343}$

∴ The required probability distribution is $\begin{pmatrix}0&1&2&3\\\frac{125}{343}&\frac{150}{343}&\frac{60}{343}&\frac{8}{343}\end{pmatrix}$