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Hey there! I'm a Year 12 Australian student and I'm currently stuck on this question.

A group of seven people is to be randomly selected from twelve people, seven men and five women. What is the probability that the group so formed contains:

A) more women than men?

B) more men than women?

Sue is one of five women.

C) what is the probability that use is one of the seven selected?

D) Given that the selected group contains more women than men, what is the probability that Sue is one of the Seven connected.

Parts A,B and C I have the answer to which are 49/198, 149/198 and 7/12 respectively. I am looking for the answer to part D, the answer is 23/28, however, I require a mathematical process.

Yes, A, B, and C are answered correctly.

For D, to get more womem, we have to look at each case individually.

To get more women, we must have 4 or 5.

When there are 4, there is a 4/5 chance of Sue being in there.

When there are 5, she is definitely there.

C(a,b) is used for "a choose b".

To get 4 women, 3 must be guys, and that C(5,4)C(7,3) = 5*35 = 175.

Since there are 175 ways of 4 women, 4/5 of that is 140.

To get 5 women, 2 must be guys, and that is C(5,5)C(7,2) = 1*21 = 21.

Since all of the women were chosen, Sue is among them.

Since there are 21 ways of getting 5 women, add that to 140 to get 161.

Now the total ways is 175+21=196.

Both of these are divisible by 7 since 7*23 = 161 and 7*28=196, so that reduces to 23/28.

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