Confused about the wording of this probability problem

[Xyius]

Homework Statement

From a drawer that contains 10 pair of gloves, six gloves are selected randomly. Let X be the number of pairs of gloves obtained. Find the probability mass function.I honestly do not know what they mean in this problem. What am I looking for probability wise? Am I looking for 6 specific gloves in a box of 20? (10 pair) If this were the case, when x=2 I could potentially have 2 probabilities to get the next glove I want depending on if I drew one on the first draw or not.

For example, If I wanted 6 gloves out of the 20, the probability of be drawing one of the 6 would be 6/20. But when I make a second draw, the probability could either be 6/19 (Meaning I didn't draw one of the gloves I wanted) or 5/19 (meaning I did). This ambiguity leads me to believe my thinking of the problem is incorrect but I honestly cannot infer any other scenario they could mean by "6 are drawn randomly."

Can anyone give me their 2 cents?

Thanks

 

[jamesrc]

I think the problem means that there are 10 unique pairs of gloves in the drawer and X is the number of pairs of gloves after randomly choosing 6 individual gloves. So one possibility is that you chose 6 gloves and none of them pair up, meaning that you have 0 pairs (X=0). Another possibility would be that you picked 3 pairs of gloves. You can come up with a probability for each possibility for X (and when you add them together, you should find that the sum of all of the probabilities is 1).

 

[Xyius]

I think the problem means that there are 10 unique pairs of gloves in the drawer and X is the number of pairs of gloves after randomly choosing 6 individual gloves. So one possibility is that you chose 6 gloves and none of them pair up, meaning that you have 0 pairs (X=0). Another possibility would be that you picked 3 pairs of gloves. You can come up with a probability for each possibility for X (and when you add them together, you should find that the sum of all of the probabilities is 1).

Yes that makes MUCH more sense! Thank you very much I believe I can now do this problem :D

 

[Xyius]

So it appears I have hit another road block here. There is still some ambiguity. Am I calculating the probability that there are paired up gloves in the 6 that I have chosen? Or the probability that there are pairs in what is left over given that I have taken out 6 random gloves? Either way, I am having trouble getting the value of the probabilities. :\

EDIT: I feel like this problem shouldn't be this difficult! I cannot seem to get any sort of probabilities.

My logic for drawing 0 pairs is, At first you have 20 choices. Then 18 since you cannot draw the same type of glove. Then 16 for the same logic. ect.
All this is divided by the total number of glove combinations which is \binom{20}{6}. This yields a number greater than 1 though!

 

[Xyius]

So I managed to find the solutions manual for this book and I do NOT understand their expression for the probability.

P(X=i)=\frac{ \binom{10}{i}\binom{10-i}{6-2i} 2^{6-2i}}{\binom{20}{6}}

None of these terms make sense to me except the denominator

 

[awkward]

Consider, for example, the probability of getting exactly 2 pairs of gloves. There are \binom{10}{6} ways to select the 6 gloves, all of which we assume are equally likely, so we need to count the number of selections which contain 2 pairs. There are \binom{10}{2} ways to select the two pairs. The only tricky part is counting the number of ways to select the remaining 2 gloves, because they must not contain a pair-- otherwise we would have 3 pairs, not 2. There are \binom{8}{2} ways to select the 2 pairs from which we draw the unmatched gloves, and given a pair, there are 2 choices for the glove; so all together there are \binom{10}{2} \binom{8}{2} 2^2 ways to select the 6 gloves containing 2 pairs.

 

[awkward]

Consider, for example, the probability of getting exactly 2 pairs of gloves. There are \binom{10}{6} ways to select the 6 gloves, all of which we assume are equally likely, so we need to count the number of selections which contain 2 pairs. There are \binom{10}{2} ways to select the two pairs. The only tricky part is counting the number of ways to select the remaining 2 gloves, because they must not contain a pair-- otherwise we would have 3 pairs, not 2. There are \binom{8}{2} ways to select the 2 pairs from which we draw the unmatched gloves, and given a pair, there are 2 choices for the glove; so all together there are \binom{10}{2} \binom{8}{2} 2^2 ways to select the 6 gloves containing 2 pairs.

Oops, I should have said there are \binom{20}{6} ways to select the 6 gloves. But I guess you knew that anyway.

 

[Xyius]

Oops, I should have said there are \binom{20}{6} ways to select the 6 gloves. But I guess you knew that anyway.

Yeah I got that ha. Thanks so much for your explanation. I forgot to thank you when I read your post initially because I was in a rush. (I was reading it on my ipod).

I am always so grateful for those who put in time to help my confusion. :D