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Binomial distribution

  1. Mar 10, 2009 #1
    I'm trying to figure out this problem but i keep getting stuck.

    The problem statement, all variables and given/known data

    A woman wants to have a 95% chance for a least a one boy and at least one girl. What is the minimum number of children that she should plan to have? Assume that the event that a child is a girl and a boy is equiprobable and independent of the gender of the other children born in the family.

    Relevant equations

    So i know you should use Pr(K = k) = (n\choose k)p^k(1-p)^(n-k)

    The attempt at a solution

    since the it's a 95% chance Pr(K=k)= .95
    probability of a boy or girl is 50% so it's .5

    .95= (n\choose k).5^k(1-.5)^(n-k)
    but then how would you solve or find what n and k is?
    Am i missing something here?

    Can anyone help! thank you in advance.
  2. jcsd
  3. Mar 10, 2009 #2
    In other words the probability of n children being either all boys or all girls is less than 5%.
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