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 the probability distribution of $X$ is given by
(1)
(2)
(3)
(4)
$\left(\frac{3}{4}\right)^2$
$\left(\frac{3}{5}\right)^2$
$\left(\frac{3}{2}\right)^2$
$\left(\frac{3}{7}\right)^3$
Solution
Probability of drawing zero black ball $={ }^3 \mathrm{C}_0\left(\frac{3}{7}\right)^0\left(\frac{4}{7}\right)^3=\left(\frac{4}{7}\right)^3$
Probability of drawing one black ball
$={ }^3 \mathrm{C}_1\left(\frac{3}{7}\right)^1\left(\frac{4}{7}\right)^2=3 \times \frac{3}{7} \times\left(\frac{4}{7}\right)^2=\frac{9}{7}\left(\frac{4}{7}\right)^2$
Probability of drawing two black balls
$={ }^3 C_2\left(\frac{3}{7}\right)^2\left(\frac{4}{7}\right)^1=3 \times\left(\frac{3}{7}\right)^2 \times \frac{4}{7}=\frac{12}{7}\left(\frac{3}{7}\right)^2$
Probability of drawing three black balls $={ }^3 \mathrm{C}_3\left(\frac{3}{7}\right)^3\left(\frac{4}{7}\right)^0=\left(\frac{3}{7}\right)^3$
(2) 
(3) 
(4) 