anonymous
  • anonymous
4 orange, 6 green, 9 blue, and 1 yellow M & M's in a bowl. What is the probability of drawing 2 orange and then 1 green M & M (without replacement)?
Mathematics
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SOLVED
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chestercat
  • chestercat
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Zarkon
  • Zarkon
if you are grabbing 3 total then \[\frac{{4 \choose 2}{6 \choose 1}}{{20 \choose 3}}\]
anonymous
  • anonymous
Zarkon... would you mind explaining please? This is my weakness.
Zarkon
  • Zarkon
\[\frac{{4 \choose 2}{6 \choose 1}}{{20 \choose 3}}=\frac{3}{95}\] \({4 \choose 2}\) is the number of ways to choose 2 orange M&M's from 4 \({6 \choose 1}\) is the number of ways to choose 1 green M&M's from 6 \({20 \choose 3}\) is the number of ways to do the experament: Choosing 3 m&m's from the 20 total

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Zarkon
  • Zarkon
here \[{n\choose k}=\, _nC_k=\,^nC_k=C_k^n=C(n,k)=\frac{n!}{k!(n-k)!}\]
anonymous
  • anonymous
Zarkon... the only part I don't understand is the 20/3 and how that equates to 95?
Zarkon
  • Zarkon
there are 20 m&m's...you are grabbing 3 \[{20\choose 3}=1140\]
anonymous
  • anonymous
Zarkon... sorry, but I still do no understand. 4/2 x 6/1 = 3. This I understand. But I don't understand how 20/3 equates to 95 or 1140. Thanks for taking the time. I appreciate it.
Zarkon
  • Zarkon
\[{20\choose 3}=\frac{20!}{3!(20-3)!}=\frac{20!}{3!17!}=\frac{20\cdot 19\cdot 18}{3\cdot 2\cdot 1}=1140\]
anonymous
  • anonymous
ok, cool. But as far as the original questions is concerned... how did you end up with 95 on the bottom?
Zarkon
  • Zarkon
\[{4\choose2}=6\] \[{6\choose1}=6\] \[{4\choose2}{6\choose1}=6\cdot 6=36\] \[\frac{36}{1140}=\frac{3}{95}\]

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