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

In summary, the probability that a particular mouse, Mike Mouse, will be included in the control group is 4/9, based on the Hypergeometric Distribution with A=1 (Mike Mouse), B=8 (Other Mice), n=4, and x=1. This was confirmed by another tutor and professor, who also used this method correctly last year for the same problem.
  • #1
redone632
14
0

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?
 
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  • #2
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.
 
  • #3
rock.freak667 said:
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?
 
  • #4
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.
 
  • #5
Dick said:
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
 
  • #6
redone632 said:
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?
 

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

1. What is a hypergeometric question?

A hypergeometric question is a type of statistical question that involves drawing a specific number of items from a set without replacement. It is commonly used to calculate the probability of obtaining a certain number of items with specific characteristics from a larger population.

2. How is a hypergeometric question different from a binomial question?

Both hypergeometric and binomial questions involve calculating probabilities, but the main difference is that a binomial question involves drawing items with replacement, while a hypergeometric question involves drawing items without replacement. This means that in a hypergeometric question, the probability of drawing an item changes with each draw, whereas in a binomial question, the probability remains constant.

3. What are some real-life applications of hypergeometric questions?

Hypergeometric questions are commonly used in genetics, epidemiology, and quality control. For example, they can be used to calculate the likelihood of obtaining a specific genetic makeup in a population, or the probability of a certain number of defective products in a batch.

4. Can hypergeometric questions be solved without using formulas?

While it is possible to solve hypergeometric questions using formulas, there are also calculators and software programs available that can be used to find solutions. Additionally, in some cases, approximations or simulations can be used to estimate the probability without using a formula.

5. What is the significance of the hypergeometric distribution?

The hypergeometric distribution is significant because it allows for the calculation of probabilities in situations where the sample is drawn without replacement. This is important in many real-life scenarios, such as medical research or quality control, where it is not feasible or ethical to replace items after they have been selected. The hypergeometric distribution also has applications in other areas of mathematics, such as combinatorics and geometry.

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