bluementor wrote:720dreaming wrote:I got A, 1/126. Here is my logic.
First girl picks a marble. Prob 1.
Second girl must pick the same color marble. Prob (4/9)
Third girl must pick the same color marble. Prob (3/8)
Fourth girl must pick the same color marble. Prob (2/7)
Fifth girl must pick the same color marble. Prob (1/6)
So....
1*(4/9)*(3/8)*(2/7)*(1/6)=1/126
I don't get it. Why do you assume that all the girls will pick the marbles first before the boys do or vice versa? If the girls and boys picked a marble each in turns, the result would be different.
The probability of a second girl to picking the same color marble can be any of the following:
1. 4/9 : if she is the 2nd person picking a marble
2. 4/8: if she is the 3rd person picking a marble (i.e. the first boy picks before her)
3. 4/7: if 2 boys pick marbles before the 2nd girl
...and so on, and a similar logic for the 3rd, 4th and 5th girl.
Could anyone please explain? Thanks.
-BM-
You're free to assume the boys and girls pick the marbles in any order you find convenient (or if you like, you can assume they all pick marbles simultaneously). Provided you do the math right, the answer will be the same in every case, which should make intuitive sense. If the boys and girls alternate, as in your scenario above, you could calculate the result as follows:
1st (girl): can pick any marble
2nd (boy): must pick the colour the first girl did not pick: 5/9
3rd (girl): must pick same as the first girl: 4/8
4th (boy): must pick same as the first boy: 4/7
5th (girl): must pick same as other girls: 3/6
and so on. Multiplying these, you get:
(5/9)*(4/8)*(4/7)*(3/6)*(3/5)*(2/4)*(2/3)*(1/2)
= (5*4*4*3*3*2*2)/9!
= (4*3*2)/(9*8*7*6)
= 1/(3*7*6)
= 1/126
I'd find this approach a lot less convenient then the approaches suggested above, but there's no reason why it shouldn't give you the same answer.
