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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?

Solution

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1 of 2Given:

$n=6$

$p=90\%=0.9$

Definition binomial probability:

$P(X=k)=_nC_k\cdot p^k\cdot (1-p)^{n-k}=\dfrac{n!}{k!(n-k)!}\cdot p^k\cdot (1-p)^{n-k}$

Evaluate at $k=0$:

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

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