debatable question
to me 1 st is suffcient, we can calculate the prob of 1 girl and 4 boys
from 2 we know that 4 boys are already in committee, but we are not given that we have at least 1 girl to chose from to me it is possible that we have no girls at all, as we are given clearly only 9 children
my pick for A

clock60 wrote:debatable question
to me 1 st is suffcient, we can calculate the prob of 1 girl and 4 boys
from 2 we know that 4 boys are already in committee, but we are not given that we have at least 1 girl to chose from to me it is possible that we have no girls at all, as we are given clearly only 9 children
my pick for A

The official answer is D

shilpaqueen wrote:A committee of 5 is to be chosen from a group of 9 children. What is the probality that there is atleast 1 girl in the commitee?
1) there are 8 boys and 1 girl in the group
2) there are 4 boys in the commitee

I am Convinced that the answer should be D, But What if have this question on the Problem Solving Section,

Guys Please Explain How to solve it,

Hi math guru's
How can statement 2 be sufficient here ?

4 boys in the commitee...
there can be 8B and 1G in the group, or 7B and 2G in the group etc...
probablity changes as no of girls changes.