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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 -

    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 -

    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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