A bag contains 4 red and 3 black balls. One ball is drawn and then replaced in the bag and the process is…
A bag contains 4 red and 3 black balls. One ball is drawn and then replaced in the bag and the process is repeated. Let X denote the number of times black ball is drawn in 3 draws. Assuming that at each draw each ball is equally likely to be selected, then probability distribution of X is given by
Solution
X denote the number of times black ball is drawn.
Since a ball is drawn three times, $0,1,2$ and 3 are possible values of X .
Probability of getting a black ball in a single draw from bag is $\mathrm{p}=\frac{3}{7}$ and $\mathrm{q}=1-\frac{3}{7}=\frac{4}{7}$
$\begin{aligned}
& \mathrm{P}[\mathrm{X}=0]=\mathrm{P}[\text { no black ball }]=\mathrm{qqq}=\mathrm{q}^3=\left(\frac{4}{7}\right)^3 \\
& \mathrm{P}[\mathrm{X}=1]=\mathrm{P}[\text { one black ball }] \\
&=\mathrm{pqq}+\mathrm{qpq}+\mathrm{qqp} \\
&=3 \mathrm{pq}^2 \\
&=3\left(\frac{3}{7}\right)\left(\frac{4}{7}\right)^2=\frac{9}{7}\left(\frac{4}{7}\right)^2 \\
& \begin{aligned}
\mathrm{P}[\mathrm{X}=2] & =\mathrm{P}[\text { two black balls }] \\
& =\mathrm{ppq}+\text { pqp }+\mathrm{qpp} \\
& =3 \mathrm{p}^2 \mathrm{q}=3\left(\frac{3}{7}\right)^2\left(\frac{4}{7}\right)=\frac{12}{7}\left(\frac{3}{7}\right)^2
\end{aligned}
\end{aligned}$
$\mathrm{P}[\mathrm{X}=3]=\mathrm{P}[\text { three black balls }]=\mathrm{ppp}=\mathrm{p}^3=\left(\frac{3}{7}\right)^3$
$\therefore \quad$ Option (A) is correct.
