Question

From $4$ red balls, $2$ white balls and $4$ black balls, four balls are selected. The probability of getting $2$ red balls is

• A. 7/21
• B. 8/21
• C. 9/21
• D. 10/21
• E. 11/21

Answer 1

Answer: (C)

We have, $4$ red, $2$ white and $4$ black balls.
$\therefore $ Total balls $=4+2+4=10$
Number of ways of selecting $4$ balls from
$10$ balls $={ }^{10} C_{4}$
$\therefore n(S)={ }^{10} C_{4}$
Let $E=$ Event getting $2$ red balls
$\therefore n(E)={ }^{4} C_{2} \times{ }^{6} C_{2}$
$\therefore $ Required probability $=p(E)$
$=\frac{n(E)}{n(S)} $
$= \frac{{ }^{4} C_{2} \times{ }^{6} C_{2}}{{ }^{10} C_{4}} $
$= \frac{\frac{4 \times 3}{2 \times 1} \times \frac{6 \times 5}{2 \times 1}}{\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}} $
$= \frac{6 \times 15}{10 \times 3 \times 7} $
$= \frac{3}{7} $
$= \frac{9}{21} $