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Math Help - Binomial Distribution Problem

  1. #1
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    Binomial Distribution Problem

    Hi, below is a problem under binomial distribution and i have answered the problem. Can someone advise if my solution to A below is correct please. Please help me also to solve probability for question 1B.

    1. Mrs. Logan has taken 8 cuttings from a plant which won her first prize in last year’s country fair. She has set the cuttings in individual pots and hopes they will take a root. On the basis of past experience, she knows there is a 0.7 probability that a given cutting will take root.
    A. Find the probability that at least 6 of the cuttings take root.
    Solution:
    Formula: P= nCr pr qn-r
    Where:
    p= 0.7
    q =0.3
    n = 8
    r = {6,7,8}
    P(X=6) = 8C6 (0.7)6 (0.3)2
    = 28 (0.7)6 (0.3)2
    = 28 (0.1176)(0.09)
    = 0.2964
    P(X=7) = 8C7 (0.7)7 (0.3)1
    = 8 (0.7)7 (0.3)1
    = 8 (0.0824)(0.3)
    = 0.1978
    P(X=8) = 8C8 (0.7)8 (0.3)0
    = 1 (0.7)8 (0.3)0
    = 1 (0.0576)(1)
    = 0.0576
    Since all events (X=6, X=7, X=8) are mutually exclusive, then
    P(X>6) = P(X=6) + P(X=7) + P(X=8)
    = 0.2964 + 0.1978 + 0.0576
    P(X>6) = 0.5518 or 55%

    B. Find the probability that more than 2 of the cuttings will take root.


    Thanks in advance.

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  2. #2
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    your answer to part A is correct


    For part B

     P(X  > 2) = 1-P(X \leq 2) = 1 - P(X=0) - P(X=1) - P(X=2)

    Alternatively you could just lookup the answer, if you are allowed to use probability tables.
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  3. #3
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    Quote Originally Posted by SpringFan25 View Post
    your answer to part A is correct


    For part B

     P(X  > 2) = 1-P(X \leq 2) = 1 - P(X=0) - P(X=1) - P(X=2)

    Alternatively you could just lookup the answer, if you are allowed to use probability tables.
    Hi SpringFan25, thank you for your advice. Given the formula you provided for part B. Below is my calculation.

    P(X=0) = 8C0 (0.7)^0 (0.3)^8
    = 8 (1)(0.000066)
    = 0.00053
    P(X=1) = 8C1 (0.7)^1 (0.3)^7
    = 8 (0.7)(0.000219)
    = 0.000123
    P(X=2) = 8C2 (0.7)^2 (0.3)^6
    = 28 (0.49)(0.000729)
    = 0.010002

    P(X>2) = 1 – P(X < 2) = 1 – P(X=0) – P(X=1) – P(X=2)
    = 1 – 0.00053 – 0.000123 – 0.010002
    = 0.9893
    P(X>2) = 1 – 0.9893 = 0.0107


    Is the above computation is correct? Kindly confirm.
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  4. #4
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    Quote Originally Posted by jovic1104 View Post
    Hi SpringFan25, thank you for your advice. Given the formula you provided for part B. Below is my calculation.

    P(X=0) = 8C0 (0.7)^0 (0.3)^8
    = 8 (1)(0.000066)
    = 0.00053
    P(X=1) = 8C1 (0.7)^1 (0.3)^7
    = 8 (0.7)(0.000219)
    = 0.000123
    P(X=2) = 8C2 (0.7)^2 (0.3)^6
    = 28 (0.49)(0.000729)
    = 0.010002

    P(X>2) = 1 – P(X < 2) = 1 – P(X=0) – P(X=1) – P(X=2)
    = 1 – 0.00053 – 0.000123 – 0.010002
    = 0.9893
    ok so far (i got a diff answer but its probably rounding error)

    P(X>2) = 1 – 0.9893 = 0.0107
    No, you already did the 1-(stuff). You dont have to do it teice.
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  5. #5
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    Quote Originally Posted by SpringFan25 View Post
    ok so far (i got a diff answer but its probably rounding error)


    No, you already did the 1-(stuff). You dont have to do it teice.
    Sorry for the error. So the final answer must be:

    P(X>2) = 0.9893

    I appriciate your help. Many thanks.
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