Math
posted by Kelsey .
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?

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 
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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