Posted by **Kelsey** on Friday, May 16, 2008 at 3:39am.

I need help with these probability problems.

1). Five chips are selected from a bag without replacement. The bag originally contained 6 yellow chips and 8 red chips. In how many ways can you choose 5 chips from the bag?

I did 14 C 5 and got 2002, is that correct.

2). In how many ways can you choose no yellow chips?

- Math -
**Reiny**, Friday, May 16, 2008 at 7:55am
no

you have to look at the results,

you have 5 chips selected, that could be one of the following cases :

0Y 5R .... RRRRR, only 1 way OR 5!/0!5!)

1Y 4R .... YRRRR, OR RYRRR, ... 5!/4! = 5

2Y 3R .... 5!/(2!3!) = 10

3Y 2R .... 5!/(3!2!) = 10

4Y 1R .... 5!/(4!1!) = 5

5Y OR .... 5!/(5!0!) = 1

total number of ways is 32

This is not a probability question, but rather based on the little formula

for the number of ways that you can arrange p things, q alike of one kind, and r alike of another kind, which is

p!/(q!r!)

2) the number of ways you can choose no yellow chips is 1, namely RRRRR

Had you asked "what is the probability of choosing no yellow chip in choosing any 5 chips from the above that would be

1/( 14C5 ) = 1/2002

- Math -
**Reiny**, Friday, May 16, 2008 at 8:12am
Correction:

I said at the end

"Had you asked "what is the probability of choosing no yellow chip in choosing any 5 chips from the above that would be

1/( 14C5 ) = 1/2002 "

that should have been 1/32

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