Question

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)?

Answer 1

if you are grabbing 3 total then

\[\frac{{4 \choose 2}{6 \choose 1}}{{20 \choose 3}}\]

Answer 2

Zarkon... would you mind explaining please? This is my weakness.

Answer 3

\[\frac{{4 \choose 2}{6 \choose 1}}{{20 \choose 3}}=\frac{3}{95}\]

\({4 \choose 2}\) 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

Answer 4

here
\[{n\choose k}=\, _nC_k=\,^nC_k=C_k^n=C(n,k)=\frac{n!}{k!(n-k)!}\]

Answer 5

Zarkon... the only part I don't understand is the 20/3 and how that equates to 95?

Answer 6

there are 20 m&m's...you are grabbing 3

\[{20\choose 3}=1140\]

Answer 7

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.

Answer 8

\[{20\choose 3}=\frac{20!}{3!(20-3)!}=\frac{20!}{3!17!}=\frac{20\cdot 19\cdot 18}{3\cdot 2\cdot 1}=1140\]

Answer 9

ok, cool. But as far as the original questions is concerned... how did you end up with 95 on the bottom?

Answer 10

\[{4\choose2}=6\]
\[{6\choose1}=6\]
\[{4\choose2}{6\choose1}=6\cdot 6=36\]

\[\frac{36}{1140}=\frac{3}{95}\]