Binomial distribution of children

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SUMMARY

The discussion centers on calculating the minimum number of children a woman should have to ensure a 95% probability of having at least one boy and one girl. The relevant equation used is the binomial probability formula: Pr(K = k) = (n choose k)p^k(1-p)^(n-k), where p is the probability of having a boy or girl, set at 0.5. The user seeks assistance in determining the values of n (total children) and k (children of one gender) to satisfy the condition that the probability of having all boys or all girls is less than 5%.

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  • Knowledge of combinatorial mathematics, particularly the binomial coefficient
  • Basic algebra skills for solving equations
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kay123
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I'm trying to figure out this problem but i keep getting stuck.

Homework Statement

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.
 
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kay123 said:
A woman wants to have a 95% chance for a least a one boy and at least one girl.

In other words the probability of n children being either all boys or all girls is less than 5%.
 

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