Find Probability of Male Mouse Given Gray: 8 Mice, 3 Males

[navi]

HELP

There are 8 mice in a cage... 3 white males, 3 gray females, and 2 gray males. Two mice are selected simultaneously and at random, and their colors are noted. Find the pr that at least one mouse is a male given that exactly ones is grey.

I am not sure if I set up the tree correctly, so I just went by with combinations. I did: C(3,1)C(2,1)/C(3,1)C(2,1)+C(3,1)C(3,1), but that did not work. Why was that approach wrong?

 

[castor28]

HELP

There are 8 mice in a cage... 3 white males, 3 gray females, and 2 gray males. Two mice are selected simultaneously and at random, and their colors are noted. Find the pr that at least one mouse is a male given that exactly ones is grey.

I am not sure if I set up the tree correctly, so I just went by with combinations. I did: C(3,1)C(2,1)/C(3,1)C(2,1)+C(3,1)C(3,1), but that did not work. Why was that approach wrong?

Hi navi,

I think you may be making this harder than it is.

If exactly one mouse is gray, the other must be white. What do you know about white mice?

 

[HOI]

As castor28 said, if there are only two colors of mice, white and gray, two are chosen, and "exactly one is gray" then the other must be white!

Perhaps you meant "at least one is gray". That's a more interesting problem! Since there are 5 gray mice and three white mice, the probability the first mouse chosen is white is 3/8, then there are 5 gray mice and two white mice. The probability the second is chosen is gray is 5/7. The probability a white and a gray mouse are chosen, in that order, is (3/8)(5/7)= 15/56.

The probability the first mouse chosen is gray is 5/8. In that case, there are 4 gray mice and 3 white mice so the probability the second mouse chosen is white is 3/7. The probability that a gray and white mouse are chosen, in that order is (5/8)(3/7)= 15/56. It should be no surprise that this is the same as before.

Now, find the probability the two mice are both gray. As before, the probability the first mouse chosen is gray is 5/8. The probability the second mouse chose is also gray is 4/7 so the probability two gray mice are chosen is (5/8)(4/7)= 20/56.

We do not need to find the probability that the two mice are both white since we have "at least one of the two mice is gray". The "measure of the sample space" is (15/56)+ (15/56)+ (20/56)= 50/56. Out of that "sample space" the probability that "the other mouse is white is (15/56+ 15/56)/(50/56)= (30/56)/(50/56)= 30/50= 3/5.