What is the probability that Mike Mouse will be included in the control group?

[redone632]

Homework Statement

Suppose 9 mice are available for a study of a possible carcinogen and 4 of them will form a control group (i.e. will not receive the substance). Assuming that a random sample of 4 mice are selected, what is the probability that a particular mouse, Mike Mouse will be included in the control group?

Homework Equations

Hypergeometric

The Attempt at a Solution

So I'm a part time tutor and a lot of the students are coming to me about this question. The way I have done it is as a Hypergeometric Distribution with A=1 (Mike Mouse) B=8 (Other Mice) n=4 x=1.

(1C1 * 8C3) / 9C4 = .4444

The students have come back and told me that it was wrong and the answer they got from the Professor was.

(4C1 * 5C3) / 9C4 = .3175

I have talked to another tutor (which is also a professor) and he did it the same way as I did. He also said that this happened last year and went to the professor last year and they said that he was right (my answer).

So now that it's come up again, I'm second guessing myself. Any input on which way is right?

 

[rock.freak667]

You'll be choosing Mike Mouse from 4, ⁴C₁. There are 5 left and you must pick the other three mice from that, so ⁵C₃.

The total number of choices is ⁹C₄.

So I think your professor is correct.

 

[redone632]

You'll be choosing Mike Mouse from 4, ⁴C₁. There are 5 left and you must pick the other three mice from that, so ⁵C₃.

The total number of choices is ⁹C₄.

So I think your professor is correct.

I don't understand why you chose 4C1 and 5C3. The way I thought of it was just like any other Hypergeometric problem. For example, you have a bag of 9 marbles, 1 black and 8 white. What is the probably that you will get the black marble when you pick 4 marbles without replacement. So there is 1 black marble in total and we want to chose 1 and there is 8 white marbles in total and we want to chose 3. So (1C1 * 8C3) / 9C4. How is this problem different from the Mike Mouse one?

 

[Dick]

It's not different. There's an even simpler way to phrase it. You are choosing 4 from a group of 9. The odds any particular one is in the group is 4/9.

 

[redone632]

It's not different. There's an even simpler way to phrase it. You are choosing 4 from a group of 9. The odds any particular one is in the group is 4/9.

Hmm, I never really thought of it like that. So my original answer was correct?
(1C1 * 8C3) / 9C4

 

[Dick]

Hmm, I never really thought of it like that. So my original answer was correct?
(1C1 * 8C3) / 9C4

Well, sure. You designate one, pick it and then pick 3 others and divide by all possible choices. That's still 4/9, right?