Q.
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)
$\frac{1}{27}$
$\frac{2}{9}$
$\frac{4}{9}$
$\frac{8}{27}$
| 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) | $\frac{1}{27}$ | $\frac{2}{9}$ | $\frac{4}{9}$ | $\frac{8}{27}$ |
Probability - Part 2
Solution:
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 $0,1,2,3$.
$P (0)={ }^{3} C _{0} q ^{3}=1 \times\left(\frac{1}{3}\right)^{3}=\frac{1}{27}$;
$P (1)={ }^{3} C _{ I } pq ^{2}=\frac{2}{9} ; P (2)={ }^{3} C _{2} P ^{2} q=\frac{4}{9}$ and
$P (3)={ }^{3} C _{3} p ^{3}=\frac{8}{27}$
$\therefore $ The required probability distribution is
X
0
1
2
3
P(X)
$\frac{1}{27}$
$\frac{2}{9}$
$\frac{4}{9}$
$\frac{8}{27}$
| X | 0 | 1 | 2 | 3 |
| P(X) | $\frac{1}{27}$ | $\frac{2}{9}$ | $\frac{4}{9}$ | $\frac{8}{27}$ |