document.write( "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. \r
\n" ); document.write( "\n" ); document.write( "all of them dislike their mother-in-law\r
\n" ); document.write( "\n" ); document.write( "none of them dislike their mother-in-law\r
\n" ); document.write( "\n" ); document.write( " at least ten of them dislike their mother-in-law\r
\n" ); document.write( "\n" ); document.write( "no more than two of them dislike their mother-in-law \n" ); document.write( "
Algebra.Com's Answer #469685 by jam007(5)\"\" \"About 
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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
\n" ); document.write( "P( dislike their mother-in-law) p = 0.90
\n" ); document.write( "n= 12
\n" ); document.write( "Therefore, the probability distribution of X is P(x)= nCxp^x(1-p^(n-x) x=0,1,. ... ,12\r
\n" ); document.write( "\n" ); document.write( "1) P(all of them dislike their mother-in-law)= P(X=12) = (12C12)(0.90)^12(0.10)^0\r
\n" ); document.write( "\n" ); document.write( " = 0.2824
\n" ); document.write( "2) P(none of them dislike their mother-in-law) =P( X=0) = 12C0(0.90)^0(0.10)^12
\n" ); document.write( " =1x10^(-12)
\n" ); document.write( "3) P(at least ten of them dislike their mother-in-law)= P(X>=1)=1-P(X=0)
\n" ); document.write( " =1-1x10^(-12)=0.9999999999999
\n" ); document.write( "4)P(no more than two of them dislike their mother-in-law) =P( X<=2)
\n" ); document.write( " =p(X=0)+P(X=1)+P(X=2)= 5.45x10^(-9) \n" ); document.write( "
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