Question

A family has 8 girls and 4 boys. A total of 2 children must be chosen to speak on the behalf of the family at a local benefit. What is the probability that 2 girls and no boys are chosen?

A. 7/55
B. 14/33
C. 12/33
D. 1/6

Answer 1

P(GG)=(8/12)(7/11)

P(GG)=56/132

P(GG)=14/33

Answer 2

Answer:  The correct option is (B) $\dfrac{14}{33}.$

Step-by-step explanation:  Given that a family has 8 girls and 4 boys. A total of 2 children must be chosen to speak on the behalf of the family at a local benefit.

We are to find the probability that 2 girls and no boys are chosen.

Total number of children in the family = 8 + 4 =12.

Let S denote the sample space of choosing 2 children from the family of 12 children and A denote the event of choosing 2 girls and no boys.

Then, according to the given information, we have

$n(S)=^{12}C_2=\dfrac{12!}{2!(12-2)!}=\dfrac{12\times11\times10!}{2\times1\times10!}=66,\\\\\\n(A)=^8C_2\times^4C_0=\dfrac{8!}{2!(8-2)!}\times1=\dfrac{8\times7\times6!}{2\times1\times6!}=28.$

Therefore, the probability of event A is given by

$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{28}{66}=\dfrac{14}{33}.$

Thus, the required probability is $\dfrac{14}{33}.$

Option (B) is CORRECT.