Question

Family A has 6 members of which 4 are males and 2 are females & family B has 5 members of both genders. (Assume it is equi-probable for a member of unknown gender to be a male or female).

Probability that family B has more females than males is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $\frac{3}{4}$

Answer 1

Answer: (A)

$\frac{1}{2}$

Family A has $6$ mambers $:4$ males and $2$ females.

Family B has $5$ members $:$

Let the probability of family B having females be $\frac{1}{2}$

In Family A, ${}^6{C}_4\times {}^6{C}_2=225$

In Family B, given that no. of females greater than males, females be $3$ and males be $2$ or females be $4$ and males be $1$

$\Rightarrow {}^5{C}_3\times {}^5{C}_2=100$ and ${}^5{C}_4\times {}^5{C}_1=25$

So, in Family B $=\frac{25}{100}\times 2=\frac{2}{4}=\frac{1}{2}.$

Hence, the answer is $\frac{1}{2}.$