Looking for confirmation on probability homework question?

[Topgun_68]

Looking for confirmation on probability!

Sorry, posted in wrong forum. It was a probability question which is why I posted it here. Can someone move it to homework section if need be. Thanks!
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Can someone let me know if I am doing this correctly, because the answer I am getting seems to small. I thought the probability would be much higher since the woman out number the men by over 50% and we are just looking for 1.

Question: Suppose that a programming team has 17 women and 15 men. Three people must be chosen to work on a special project. What is the probability that the group selected has at least 1 women in it?

My work:

(17C1) -> To choose 1 women
(15C2) -> To choose 2 men
(32C3) -> Total groups of 3 among all people

(17C1)(15C2)
--------------
32C3 17 * 105
--------
4960

1785
----- = .360
4950 I've seem some examples where they subtract this answer from 1 which not looks more reasonable, but is it correct and can someone explain why?

ex.. 1 - .360 = .640

Thanks for any input!

 

[DrClaude]

Read the statement carefully: at least one woman.

 

[Topgun_68]

Ahh, so would I use:

(17C1 + 17C2 + 17C3) / 32C3 to account for having 1,2 or 3 women chosen? Or would I have to account for the 8 possibilities of women/men?

Thanks for feedback. I hope this doesn't get deleted for my lack of reading. I was just getting ready to erase it and repost..

Read the statement carefully: at least one woman.

 

[DrClaude]

(17C1 + 17C2 + 17C3) / 32C3 to account for having 1,2 or 3 women chosen?

Not quite. If you have for instance exactly 2 women, is the number of possiblities 17C2? Aren't you forgetting something?

And this might be a good time to use the other method you mention: 1 - something.

 

[Topgun_68]

I'm forgetting the men.. So would it be..

(17C1)(15C2) + (17C2)(15C1) + 17C3
-------------------------------------
32C3

I was thinking of subtracting the result from 1. Do I have to account for no women..

Not quite. If you have for instance exactly 2 women, is the number of possiblities 17C2? Aren't you forgetting something?

And this might be a good time to use the other method you mention: 1 - something.

 

[DrClaude]

I'm forgetting the men.. So would it be..

(17C1)(15C2) + (17C2)(15C1) + 17C3
-------------------------------------
32C3

Right.

I was thinking of subtracting the result from 1.

Why would you do that? Calculate the value numerically and see what makes sense. There is a way to calculate the initial problem as 1-something. Can you figure that one out?

Do I have to account for no women..

Account in what way?

 

[Topgun_68]

Hmm, I think I got it now.

1 - ((15C3) / (32C3)) = .908

So calculate the probability of no women from the total, than subtract it from 1.

I calculated it my long way above and I get the same answer. Can you confirm this is correct :<)

Thanks for your help on this.

Right.

Right.Why would you do that? Calculate the value numerically and see what makes sense. There is a way to calculate the initial problem as 1-something. Can you figure that one out?Account in what way?

 

[DrClaude]

Hmm, I think I got it now.

1 - ((15C3) / (32C3)) = .908

So calculate the probability of no women from the total, than subtract it from 1.

I calculated it my long way above and I get the same answer. Can you confirm this is correct :<)

That is indeed correct.

Thanks for your help on this.

You're welcome!