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Math Help - Binomial Distribution...

  1. #1
    Member looi76's Avatar
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    Binomial Distribution...

    Question:
    Given that Y \sim B(7,\frac{2}{3}), calculate P(Y=4).

    Attempt:
    n = 7, p = \frac{2}{3}, q = 1-\frac{2}{3} = \frac{1}{3}

    P(Y=y) = ^nC_y\times p^n\times q^{n-y}

    P(Y=4) = ^7C_4\times \frac{2}{3}^4\times \frac{1}{3}^{7-4}
    <br />
P(Y=4) = 0.032

    Where did I go wrong?
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  2. #2
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    Try using brackets on your fractions.


    P(Y=4) = ^7C_4\times (\frac{2}{3})^4\times (\frac{1}{3})^{7-4}=0.256

    Is that the answer you expected?

    Actually that doesn't acount for your answer.
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  3. #3
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    mr fantastic's Avatar
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    Quote Originally Posted by a tutor View Post
    Try using brackets on your fractions.


    P(Y=4) = ^7C_4\times (\frac{2}{3})^4\times (\frac{1}{3})^{7-4}=0.256

    Is that the answer you expected?

    Actually that doesn't acount for your answer.
    And don't skimp on the ink - use BIG brackets:

    P(Y=4) = ^7C_4\times \left( \frac{2}{3} \right)^4 \times \left( \frac{1}{3} \right)^{7-4}=0.256
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