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Question

Sociologists say that 90% of married women claim that their husband's mother is the biggest bone of contention in their marriages (sex and money are lower-rated areas of contention). Suppose that six married women are having coffee together one morning. What is the probability that none of them dislike their mother-in-law?

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Given:

n=6n=6

p=90%=0.9p=90\%=0.9

Definition binomial probability:

P(X=k)=nCkpk(1p)nk=n!k!(nk)!pk(1p)nkP(X=k)=_nC_k\cdot p^k\cdot (1-p)^{n-k}=\dfrac{n!}{k!(n-k)!}\cdot p^k\cdot (1-p)^{n-k}

Evaluate at k=0k=0:

P(X=0)=6!0!(60)!0.90(10.9)60=0.000001=0.0001%P(X=0)=\dfrac{6!}{0!(6-0)!}\cdot 0.9^0\cdot (1-0.9)^{6-0}=0.000001=0.0001\%

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