Question

One day, eleven babies are born at a hospital. assuming each baby has an equal chance of being a boy or girl, what is the probability that at most nine of the eleven babies are girls?

Answers

  1. The probability of having, at most, 9 girls, is 0.9515

    How to get the probability?

    The probability that a random baby is a girl is:
    p = 0.5
    And the probability that a random baby is a boy is:
    q = 0.5
    Then the probability that, at most, 9 out of 11 babys are girls, is given by:
    1 – p(10) – p(11)
    Where P(10) is the probability that 10 of the babies are girls and p(11) is the probability that the 11 babies are girls.
    p(10) = C(11, 10)*(0.5)^10*(0.5)^1 = C(11, 9)*(0.5)^11
    Where C(11, 10) is the combinations of 10 elements that we can make with a set of 11 elements, such that:
    C(11, 10)=  11!/(11 – 10)!*10! = 11
    Replacing that, we get:
    P = 11*(0.5)^11 = 0.0054
    p(11) = C(11, 11)*0.5^11 = 1*0.5^11 = 0.0005
    Then the probability is:
    P = 1 – 0.0054 – 0.0005 = 0.9515
    The probability of having, at most, 9 girls, is 0.9515
    If you want to learn more about probability:
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