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    • Thread Starter


    Given that there are equal numbers of boys and girls born, find the probability that a family of six children which has been chosen at random has:

    (a) more girls than boys

    For this I understood that there'd have to be 4 or 5 or 6 girls to be more girls than boys. So P(X > 3) = P(X=4) + P(X=5) + P(X=6).

    So, starting with the 1st probability: P(X=4) = 6! over 5! 1! 0.5^4 x 0.5^1

    Is this the correct method. Am I meant to do this for each probablity and then add the final answers up?

    Thank you

    You could also use the fact that P(X > 3) = 1 - P(X < 3)
    • Thread Starter

    Could you show me the working out for that method ?
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Updated: May 22, 2010

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