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 $=p(E)$
$=\frac{n(E)}{n(S)}$
$=\frac{{}^4{C}_2\times {}^6{C}_2}{{}^{10}{C}_4}$
$=\frac{\frac{4\times 3}{2\times 1}\times \frac{6\times 5}{2\times 1}}{\frac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}}$
$=\frac{6\times 15}{10\times 3\times 7}$
$=\frac{3}{7}$
$=\frac{9}{21}$