Question

A bag contains 3 white, 3 black and 2 red balls. One by one, three balls are drawn without replacing them. The probability that the third ball is red, is given by:

• A. $\frac{5}{25}$
• B. $\frac{1}{12}$
• C. $\frac{1}{4}$
• D. $None$

Answer 1

The correct option is C $\frac{1}{4}$

White balls = 3 , black = 3, Red = 2.
Since drawn balls are not replaced, for third ball to be red, we have the following patterns:
$E_1=WWR,BBR,E_2=BWR,WBR;$ and
$E_3=RBR,RWR,BRR,WRR.$
$P\left(E_1\right)=2\times \frac{3}{8}\times \frac{2}{7}\times \frac{2}{6}=\frac{24}{8\times 7\times 6}P\left(E_2\right)=\frac{2\times 3\times 3\times 2}{\mathrm{8.7.6}}=\frac{36}{\mathrm{8.7.6}}P\left(E_3\right)=\frac{4\times 2\times 3\times 1}{\mathrm{8.7.6}}=\frac{24}{\mathrm{8.7.6}}$
Required probability
$=P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)=\frac{84}{\mathrm{8.7.6}}=\frac{1}{4}$