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A population consists of 25 men and 25 women. A simple random sample (draws at random without replacement) of 4 people is chosen. Find the chance that in the sample all the people are of the same gender?

Two events are dependent A and B. Multiplication Rule P(A and B)= P(A)*P(B\A)= if A and B are dependent then P(A and B)= P(A)P(B)=(4/25)(1)=?

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    $\begingroup$ Please show us your thought on the question. $\endgroup$ – mez Apr 29 '13 at 9:36
  • $\begingroup$ What has the second paragraph to do with the title and the first paragraph? $\endgroup$ – Did Apr 29 '13 at 10:30
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If I understood correctly, without replacement means that after choosing someone, you cannot choose them again, so I will solve according to this.

Let's do this in steps.

  1. We pick the first person and take them out of the room. They can be of any gender.

  2. We pick the second person. They must be of the same gender of the first person. Since the first person is no longer in the room, there are now 49 people in the room, with 24 people of the first person's gender, such that the probability that the second person is of the same gender as the first is $ \frac{24}{49} $

  3. We pick the third person, out of the room of now 48 people. There are now only 23 people of the same gender as the first person in the room, such that the probability that the third person is of the same gender as the first and second is $ \frac{23}{48}$.

  4. Now, we pick the fourth person of a room of 47 people, which contains 22 people of the same gender as the first person, such that the probability that the fourth person will be of the same gender as the first person is $ \frac{22}{47}$.

Now, we want all of these events (meaning 1,2,3,4) to occur, and therefore, we need to multiply these probabilities, in order to get that the probability that all of the people are of the same gender is $ \frac{24\cdot 23 \cdot 22}{49\cdot 48 \cdot 47}$.

In general, using this algorithm, we can find a formula for the probability to choose k people of the same gender from a room of n men and n women.

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  • $\begingroup$ Thanks Noy Soffer clearly Understood $\endgroup$ – juniorgraduation2013 Apr 29 '13 at 10:36
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The probability that the second person's gender is the same as the first's is $24/49$ (there are 24 unchosen people of that gender left and 49 unchosen people in total). The probability that the third person's gender is the same again is $23/48$, and the probability for the fourth is $22/47$.

$$(24\times23\times22)/(49\times48\times47)=\frac{253}{2303}\approx0.11.$$

More generally, if there are $n$ people of whom $n/2$ are males the equation would be $$\frac{\left(\frac{n}{2}-1\right) \left(\frac{n}{2}-2\right) \left(\frac{n}{2}-3\right)}{(n-1) (n-2) (n-3)},$$ which looks like this:

enter image description here

Note that it converges on $0.125$ as $n$ grows large because if there are very many people the proportion of men and women remains roughly equal even as a few people of one gender are taken away. Thus the probability converges to $0.5\times0.5\times0.5=0.125$. Edit: Noy Soffer beat me to it!

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