Question 770626: 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 twelve married women are having coffee together one morning. Find the following probabilities.
all of them dislike their mother-in-law
none of them dislike their mother-in-law
at least ten of them dislike their mother-in-law
no more than two of them dislike their mother-in-law
Answer by jam007(5)   (Show Source):
You can put this solution on YOUR website! The number of women out of 12 selected dislike their mother-in-law is a random variable that can be represented by X. Then X is a Binomial distribution with
P( dislike their mother-in-law) p = 0.90
n= 12
Therefore, the probability distribution of X is P(x)= nCxp^x(1-p^(n-x) x=0,1,. ... ,12
1) P(all of them dislike their mother-in-law)= P(X=12) = (12C12)(0.90)^12(0.10)^0
= 0.2824
2) P(none of them dislike their mother-in-law) =P( X=0) = 12C0(0.90)^0(0.10)^12
=1x10^(-12)
3) P(at least ten of them dislike their mother-in-law)= P(X>=1)=1-P(X=0)
=1-1x10^(-12)=0.9999999999999
4)P(no more than two of them dislike their mother-in-law) =P( X<=2)
=p(X=0)+P(X=1)+P(X=2)= 5.45x10^(-9)
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