Question

A box contains $7$ white, $5$ red and $3$ black balls. Three balls are picked randomly one at a time, and are being replaced. What is the probability of picking a white then red then a black ball?

• A. $\frac{7}{15}$
• B. $\frac{7}{225}$
• C. $\frac{1}{3}$
• D. $\frac{14}{225}$

Answer 1

The correct option is B $\frac{7}{225}$

In the box,
number of white balls= $7$
number of red balls= $5$
number of black balls= $3$
$\therefore$ Total number of balls in the box= $7+5+3=15$

To calculate the probability of compound event, first we need to calculate the probabilities of included single events.

$\therefore \text{P(white)}=\frac{\text{Number of white balls}}{\text{Total number of balls}}$

$\Rightarrow \text{P(white)}=\frac{7}{15}$

Similarly,
$\text{P(red)}=\frac{5}{15}=\frac{1}{3}$, and $\text{P(black)}=\frac{3}{15}=\frac{1}{5}$

Now,
$\text{P(white, then red, then black)=P(white)}\times \text{P(red)times P(black)}$

$\Rightarrow \text{P(white, then red, then black)}=\frac{7}{15}\times \frac{1}{3}\times \frac{1}{5}$

$\Rightarrow \text{P(white, then red, then black)}=\frac{7\times 1\times 1}{15\times 3\times 5}$

$\Rightarrow \text{P(white, then red, then black)}=\frac{7}{15\times 15}$

$\Rightarrow \text{P(white, then red, then black)}=\frac{7}{225}$

$\therefore$ The probability of picking a white then red then a black ball is $\frac{7}{225}$.