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Practice Question #41

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  • Practice Question #41

    Does anyone know how to do #41 and possibly explain it on here. I don't really 'like' the way the solution manual explained it... since that's not the way we learned in class?

    This is what I did... (but got the wrong answer)

    Add the probability of 9 people staying with the probability of 10... then multiply that by 2 since the two groups are independent.

    and I got 2*[ (10 nCr 9)*0.2*(0.8^9) + (10 nCr 10)*(0.8^10) ]

    then I subtracted the probability of BOTH groups getting 9 or 10....

    probability of both groups (group A, group B) getting 9 or 10 = P(A=9 and B=9) + P(A=9 and B=10) + P(A=10 and B=9) + P(A=10 and B=10)

    in any case, the final answer was wrong.
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