Question

Seven balls are drawn simultaneously from a bag containing $5$ white and $6$ green balls. The probability of drawing $3$ white and $4$ green balls is

• A. $\frac{7}{{}^{11}C_7}$
• B. $\frac{{}^{5}C_3+{}^{6}C_4}{{}^{11}C_7}$
• C. $\frac{{}^{5}C_2·{}^{6}C_2}{{}^{11}C_7}$
• D. $\frac{{}^{6}C_3+{}^{5}C_4}{{}^{11}C_7}$

Answer 1

The correct option is C

$\frac{{}^{5}C_2·{}^{6}C_2}{{}^{11}C_7}$

Explanation for the correct option:

Finding the required probability in terms of combinations:

The total number of balls $=11$

Therefore, the total number of cases $={}^{11}C_7$

The favourable cases ${}^{5}C_3·{}^{6}C_4={}^{5}C_2·{}^{6}C_2$

The Required Probability is $\frac{{}^{5}C_2·{}^{6}C_2}{{}^{11}C_7}\left[\because \mathrm{Probability}=\frac{\mathrm{Favourable}\mathrm{cases}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{cases}}\right]$

Therefore, the correct answer is option (C).