Question

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

X	0	1	2	3
P(X)	$\frac{4}{9}$	$\frac{5}{9}$	$\frac{7}{9}$	$\frac{1}{9}$
X	0	1	2	3
P(X)	$\frac{5}{54}$	$\frac{7}{54}$	$\frac{2}{27}$	$\frac{7}{27}$
X	0	1	2	3
P(X)	$\frac{1}{3}$	$\frac{2}{3}$	$\frac{9}{26}$	$\frac{3}{26}$
X	0	1	2	3
P(X)	$\frac{1}{27}$	$\frac{2}{9}$	$\frac{4}{9}$	$\frac{8}{27}$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (D)

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 $=5+10=15$, out of which $10$ are good. If $p=$ probability of drawing a good egg, then
$p=\frac{10}{15}=\frac{2}{3}$,
$\therefore q=1-\frac{2}{3}=\frac{1}{3}$
Thus, we have a binomial distribution with $p=\frac{2}{3},q=\frac{1}{3}$ and $n=3$.
If $X$ denotes the number of good eggs drawn, then $X$ can take values $0,1,2,3$.
$P(0)={}^3{C}_0{q}^3=1\times {\left(\frac{1}{3}\right)}^3=\frac{1}{27}$;
$P(1)={}^3{C}_{I}p{q}^2=\frac{2}{9};P(2)={}^3{C}_2{P}^{2}q=\frac{4}{9}$ and
$P(3)={}^3{C}_3{p}^3=\frac{8}{27}$
$\therefore$ The required probability distribution is

X	0	1	2	3
P(X)	$\frac{1}{27}$	$\frac{2}{9}$	$\frac{4}{9}$	$\frac{8}{27}$