From a committee of 6 girls and 4 boys, a name is selected. Then another name is selected. What is the probability that both people drawn will be girls? (1 point)

nine over twenty-five
two-fifteenths
three-fifths
one-third

Question

From a committee of 6 girls and 4 boys, a name is selected. Then another name is selected. What is the probability that both people drawn will be girls?  (1 point)

    nine over twenty-five
    two-fifteenths
    three-fifths
    one-third

anonymous · Answer

Similar to last problem:

(6/10) times (5/9) = 1/3

This is a "non-replacement" problem.  The total probability is:

P(1st is a girl) times P(2nd is a girl)

And since one girl is already picked and therefore unavailable, ther are only 5 girls and 9 total people left for the second pick.

nathan917 · Answer

I Got it. 

probability of drawing a girl in the first pick x probability of drawing a girl in the 2nd pick = 6/10 x 5/9 = (6x5) / (10x9) = 30 / 90 = 1 / 3

anonymous · Answer

You are understanding this well now!

nathan917 · Answer

Yes.

anonymous · Answer

You basically did what I did and modeled after the previous problem.  Good job, indeed!

anonymous · Answer

6c1/10c1 * 5c1/9c1 = 1/3