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Math Help - Second question

  1. #1
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    Second question

    The computer manufacturer plans to test 4 computers out of the shipment of 22 for testing. Suppose 2 of the 22 are faulty.

    a. How many ways can he choose the four to test?

    b. What are the chances that both of flawed computers are in the test group.
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  2. #2
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    Quote Originally Posted by BaBrown1s View Post
    The computer manufacturer plans to test 4 computers out of the shipment of 22 for testing. Suppose 2 of the 22 are faulty.

    a. How many ways can he choose the four to test?

    b. What are the chances that both of flawed computers are in the test group.
    a) \binom{22}{4}= 7315

    b) use binomial theorem

    P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}


    where n = 4, p= 2/22, k=2


    P(X=2) = \binom{4}{2}(\tfrac{2}{22})^2(1-\tfrac{2}{22})^{4-2}
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