Mother In Law Statistics Problem

Question

Sociologists say that 80% 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 seven married women are having coffee together one morning. Find the following probabilities. (Round your answers to three decimal places.)

(a) All of them dislike their mother-in-law. 

 

(b) None of them dislike their mother-in-law. 

 

(c) At least five of them dislike their mother-in-law. 

 

(d) No more than four of them dislike their mother-in-law. 

Answer 1

This is a binomial distribution problem with a bit of calculation involved.

Recall the binomial distribution gives the probability of r successes in n trials of independent events as:

Pr ( r successes in n trials ) = nCr x p^r x (1 - p)^(n-r) where nCr is the number of combinations of n items taken r at a time, p = the probability of success and (1-p) = the probability of failure.

Here will define p as the probability of disliking your mother-in-law = .8 and (1-p) = probability you like her

a) All of them dislike her. Means r = 7

Pr ( r =7 ) = 7C7 x .8⁷ x .2⁰ = 1 x .2097 x 1 = .210

b) None of them dislike their mother-in-law Means r = 0

Pr ( r = 0 ) = 7C0 x .8⁰ x .2⁷ = 1 x 1 x .0000128 = .000 practically 0

c) At least 5 of them dislike their mother-in-law Means r = 5 or 6 or 7 We've done r = 7

Pr ( r = 5 ) = 7C7 x .8⁵ x .2² = 21 x .32768 x .04 = .275 and

Pr ( r = 6 ) = 7C6 x .8⁶ x .2¹ = 7 x .0131072 x .2 = .018

So we have Pr ( r = 5 or 6 or 7 ) = .018 + .275 + .210 = .503 or approximately 50%

d) No more than 4 of them dislike their mother-in-law Means r = 0,1,2,3 or 4

But this is the same as 1 - Pr ( r = 5,6 or 7 ) = 1 - .503 =.497