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Here is the problem:

You have a book collection that consists of 20 horror novels, 15 romance novels, and 25 mystery novels. You randomly pick 4 books to read during a long trip. What is the probability that you pick at least one book of each type?

Please tell me if I did this right:

I got 1,500/1,711 (simplified from 472,500/487,635).

  1. I got 487,635 by finding the total number of combinations.
  2. I found the total # of combinations by doing 38C4. (38!/34!*4!=487,635)
  3. I got 472,500 by finding the number of combinations of 4 books where there is at least one of each book.
  4. I did this by adding (20C1*15C1*25C2)+(20C1*15C2*25C1)+(20C2*15C1*25C1).
  5. This simplifies to (20*15*25*24)+(20*15*14*25)+(20*19*15*25) = 180,000+142,500+105,000 = 427,500.

Thank you.

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First, in step $2$, the total number of books is $60$, so the total number of combinations is $_{60}C_4$ (although it seems your numerical answer is correct here). Then, in step $5$, in computing $_nC_2$, you've forgotten to divide by $2$. Hence, your probability should be half of what you've written. –  Jared May 8 '13 at 0:46
    
Thank you so much for your help. I got 38 from a different problem because I was copying this from my math worksheet. I see my mistake in calculating nC2. Thank you again. –  Sabrina May 8 '13 at 0:58
    
For step 2 you calculated $C^{60}_{56}$ not $C^{60}_{4}$ - They are the same number because a combination is $\frac{n!}{k!(n-k)!}$ –  Dale M May 8 '13 at 1:22
    
If we think about it, probability one of each type $1500/1711$ seems awfully high. –  André Nicolas May 8 '13 at 1:46
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1 Answer 1

What you have here is the Multivariate Hypergeometric Distribution

For your question you are interested in selecting 4 books with at least 1 from each of 3 categories. You are on the right track.

$$\begin{align} P(H\ge1,R\ge1,M\ge1)&=P(H=2,R=1,M=1)+P(H=1,R=2,M=1)+P(H=1,R=1,M=2)\\ &= \frac{\binom{20}{2}\binom{15}{1}\binom{25}{1}}{\binom{60}{4}} + \frac{\binom{20}{1}\binom{15}{2}\binom{25}{1}}{\binom{60}{4}} +\frac{\binom{20}{1}\binom{15}{1}\binom{25}{2}}{\binom{60}{4}}\\ &=\frac{\frac{20!}{2!18!}\frac{15!}{1!14!}\frac{25!}{1!24!}+\frac{20!}{1!19!}\frac{15!}{2!13!}\frac{25!}{1!24!}+\frac{20!}{1!19!}\frac{15!}{1!14!}\frac{25!}{2!23!}}{\frac{60!}{4!56!}}\\ &=\frac{\frac{20\times19\times15\times25}{2}+\frac{20\times15\times14\times25}{2}+\frac{20\times15\times25\times24}{2}}{\frac{60\times59\times58\times57}{4\times3\times2}}\\ &=\frac{71,250+52,500+90,000}{487,635}\\ &=\frac{213,750}{487,635}\\ &\approx0.438 \end{align}$$

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