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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: a)the fourth person is a woman b)the third person is a woman, given that the first person and fourth person are both men

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Imagine drawing people, one after the other, until we have drawn all $50$. Then all orderings of the people are equally likely.

So the probability the fourth person is a woman is the same as the probability that the first person is a woman, which is $\frac{25}{50}$.

Similarly, the probability that the third person is a woman, given that the second and fourth are men, is the same as the probability the third is a woman, given the first two are men. This is $\frac{25}{48}$.

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+1 Nice approach. Mine would have been woefully computational. – copper.hat Apr 30 '13 at 2:08
@copper.hat: But it is worthwhile to do the computation, and after a certain amount of work reach $1/2$. I think that was the pedagogic point Alex is making. – André Nicolas Apr 30 '13 at 4:00
Well, the computation supplies confirmation, but no intuition. Viewing the sample space as all permutations simplifies and elucidates at the same time. – copper.hat Apr 30 '13 at 5:11

Denote $0$ for man and $1$ for woman. $S$ is an event that 4th person is a woman. Obviously 8 cases for the first 3 tosses are possible: $\{000,001, \cdots ,111\}$. Hence you have to use the law of total probability: $$ P(S)= P(S|000)P(000) + P(S|001)P(001)+ \cdots +P(S|111)P(111) $$

The answer to the second question should follow easily

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