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Math Help - Bernoulli Family

  1. #1
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    Lightbulb Bernoulli Family

    A family has 4 children. Assuming that boys and girls are equally likely, calculate the probability, using fractions, and Bernoulli's formula (please use this specific formula when necessary)

    Probability that:
    (i) there are exactly 2 boys.
    (ii) there are exactly 3 girls.
    (iii) there is at least one boy.

    As far as I know, 'k' means the number of successes, 'n' is the number of Bernoulli trials, while P is the probability, hope i am right.

    Any input will do >_< Thanks...
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  2. #2
    Senior Member Dinkydoe's Avatar
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    Yes, you're right.

    If we denote the birth of a girl as a succes, that is we let X denote the number of girls,...then P(X=k)= {n \choose k}p^k(1-p)^{n-k} where p= \frac{1}{2}. (In this case the formula is ofcourse the same for the number of boys)

    Thus calculate:

    (i) P(X=2)
    (ii) P(X=3)
    (iii) P(X\geq 1) = 1- P(X< 1) = 1- P(X=0) (why?)
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  3. #3
    MHF Contributor
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    Quote Originally Posted by Aeonitis View Post
    A family has 4 children. Assuming that boys and girls are equally likely, calculate the probability, using fractions, and Bernoulli's formula (please use this specific formula when necessary)

    Probability that:
    (i) there are exactly 2 boys.
    (ii) there are exactly 3 girls.
    (iii) there is at least one boy.

    As far as I know, 'k' means the number of successes, 'n' is the number of Bernoulli trials, while P is the probability, hope i am right.

    Any input will do >_< Thanks...
    Let x be the random variable of the number of boys in the family,

    X-B(4,0.5)

    (1) Calculate P(X=2)

    (2) Calculate P(X=1)

    (3) Calculate 1-P(X=0)
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