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Math Help - Binomial ,with replacement

  1. #1
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    Binomial ,with replacement

    found this question on a website , it gives the answer as 0.27
    but it doesn't show how the author arrived at the answer
    ive never dealed with replacement before


    A bag contains 5 yellow beads and 5 blue beads (all identical apart from the colour).
    7 beads are drawn randomly from the bag, each being replaced before the next is drawn.
    Calculate the probability that
    Exactly 4 of the beads drawn are yellow

    7! / 4! 3! = 35 * multiplied by 0.5 seven times= 0.2734

    I dont know if the formula used is correct
    or i simply got lucky and matched authors answer ?
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  2. #2
    MHF Contributor harish21's Avatar
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    I would like to know why you multiplied 0.5 seven times.

    did you arrive to your answer by using this formula:

    \dbinom{n}{x}p^{x}(1-p)^{n-x}??
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  3. #3
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    I would like to know why you multiplied 0.5 seven times.

    because the chance of drawing a yellow bead or blue is 1 in 2

    did you arrive to your answer by using this formula:
    that formula you used is a bit advanced for me!!
    im a standard calculator guy !!
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  4. #4
    MHF Contributor harish21's Avatar
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    okay. let me ask you something before I answer your question.

    what would you have done if the question was same except that you had 6 yellow and 4 blue beads?
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  5. #5
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    so the new question would be
    A bag contains 6 yellow beads and 4 blue beads (all identical apart from the colour).
    7 beads are drawn randomly from the bag, each being replaced before the next is drawn.
    Calculate the probability that
    Exactly 4 of the beads drawn are yellow

    7! / 4! 3! * 0.6 *0.6 *0.6 * 0.6 * 0.4 * 0.4 * 0.4 = 0.2903
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  6. #6
    MHF Contributor harish21's Avatar
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    Right.

    So your answers are correct in both the cases. And the formula you are using is correct indeed!


    Note that you are using the same formula I have written in post #2.

    Here's an explanation(your original question):

    since there are 5 yellow and 5 blue beads,

    the probability of obtaining a yellow bead p=5/10=0.5

    the probability of obtaining a yellow bead (1-p)=1-0.5=0.5

    Total number of balls = 7.

    Let X be the number of yellow balls drawn. then,

    P(X=x)=\dbinom{n}{x}p^{x}(1-p)^{n-x} \mbox{where}\dbinom{n}{x}=\dfrac{n!}{x!(n-x)!}

    the probability of 4 yellow balls being drawn is:

    P(X=4)=\dbinom{7}{4}\times(0.5)^{4}\times(0.5)^{7-4}

    P(X=4)=\dfrac{7!}{4!\times 3!} \times (0.5)^4\times(0.5)^3

    =\dfrac{7!}{4!\times 3!} \times (0.5)^7= 0.27
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  7. #7
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    tx , found the question on a website was trying for 2 hours to find a formula to get the same answer as the
    author !!
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