# Math Help - Probability problem

1. ## Probability problem

Yet another brain freeze. "A student is munching on a bag of plain M&Ms. With 22 pieces left, 5 of which are red and 3 of which are yellow, she spills 10 pieces.
a. Find the probability that the bag still contains the 5 red and 3 yellow pieces.
b. Determine the probability that exactly 1 of each color red and yellow remain in the bag."

For a, the answer is simply $\frac{7}{11}$ or .636, correct?

I can't think of how to approach b.

2. For part b.

We must choose 4 of the 14 'other' colors, 4 of the 5 reds and 2 of the 3 yellows out of 10 chosen from 22.

I would think:

$\frac{C(14,4)C(5,4)C(3,2)}{C(22,10)}=\frac{15}{646 }\approx{0.0232...}$

That seems rather low, doesn't it?. I may be off base for whatever reason.

Soroban, Plato, etc. may be along to either concur or prove me wrong.

3. Part a asks for the probability that none of the red or yellow M&M’s are spilled.
Is that $\frac{C(14,10)}{C(22,10)}$?

For part b: $\frac{C(14,4)C(5,4)C(3,2)}{C(22,10)}$

4. For guys with all the answers, you sure pose a lot of questions

5. Apparently, I was correct. I thought it seemed right, but the probability, intuitively, seemed awful low. Oh well, they can be tricky.

The first one has a very low probability. One would think it'd be higher, huh?.

There's another way to look at part a.

$\Pi_{k=0}^{9}\left(\frac{14-k}{22-k}\right)$

part b:

$\frac{10!}{4!4!2!}\Pi_{k=0}^{3}\frac{5-k}{22-k}\cdot\Pi_{k=0}^{1}\frac{3-k}{18-k}\cdot\Pi_{k=0}^{3}\frac{14-k}{16-k}=\frac{15}{646}$

6. Originally Posted by galactus
For part b.

$\frac{C(14,4)C(5,4)C(3,2)}{C(22,10)}=\frac{15}{646 }\approx{0.0232...}$
Just curious - what is your technique for simplifying this fraction? In particular, how did you know that you could cancel out the C(14,4) from both parts?

7. $\frac{{C(14,4)C(5,4)C(3,2)}}{{C(22,10)}} = \frac{{\frac{{14!}}{{4!10!}}\frac{{5!}}{{4!1!}}\fr ac{{3!}}{{2!1!}}}}{{\frac{{22!}}{{10!12!}}}}
$

8. Actually, I ran it through my calculator to find the answer. I didn't do it by hand. But if you want to, use $\frac{n!}{(n-r)!r!}$

For C(14,4), you have $\frac{14!}{10!4!}=$

$\frac{14*13*12*11*10*9*8*7*6*5*4*3*2}{10*9*8*7*6*5 *4*3*2*4*3*2}=$

Do the cancellations and you get:

$\frac{14*13*12*11}{4*3*2*1}=1001$

Is that what you mean?.

9. Originally Posted by Plato
$\frac{{C(14,4)C(5,4)C(3,2)}}{{C(22,10)}} = \frac{{\frac{{14!}}{{4!10!}}\frac{{5!}}{{4!1!}}\fr ac{{3!}}{{2!1!}}}}{{\frac{{22!}}{{10!12!}}}}
$
yes, I understand that's the way to do it... I was just wondering if there's a simple way to know that C(14,4) or 1001 cancels out from the fraction...

10. Originally Posted by earachefl
I was just wondering if there's a simple way to know that C(14,4) or 1001 cancels out from the fraction...
I think that one just has to to the work.

11. Originally Posted by Plato
I think that one just has to to the work.
Bah, humbug!!!

12. Did you take note of the other way I showed?. Do you follow it?. Just another way.

13. Originally Posted by galactus
Did you take note of the other way I showed?. Do you follow it?. Just another way.
well, I saw it, but I don't understand it

14. Capital Pi stands for product, just like capital sigma stands for sum.

the $\frac{10!}{4!4!2!}=3150$ is the number of ways to arrange the 10 m&m's. Because there are 10 total with 4 red, 2 yellow and 4 others.

The probability of 4 out of 5 reds is $\Pi_{k=0}^{3}\frac{5-k}{18-k}=
(5/22)(4/21)(3/20)(2/19)=\frac{1}{1463}$

The probability of choosing the 2 yellows is $\Pi_{k=0}^{1}\frac{3-k}{18-k}=(3/18)(2/17)=\frac{1}{51}$

The probability of choosing the 4 others is: $\Pi_{k=0}^{3}\frac{14-k}{16-k}=(14/16)(13/15)(12/14)(11/13)=\frac{11}{20}$

3150(1/1463)(1/51)(11/20)=15/646

15. well, I'm gonna chew on that one for a while....