Sociologists say that 90% of married women claim that their husband’s mother is the biggest bone of contention in their marriages.
Suppose that six married women are having coffee together one morning. What is the probability that:
a) all of them dislike their mother-in law?
b) none of them dislike their mother-in law?
c) at least four of them dislike their mother-in law?
d) no more than three of them dislike their mother-in law?
It is a binomial distribution probability problem.
To facilitate my calculations, I will use an appropriate online (free of charge) calculator
at this web-site
https://stattrek.com/online-calculator/binomial.aspx
It provides a convenient input (and output) for all relevant options/cases.
(a) all of them dislike their mother-in law?
- number of trials n = 6;
- number of success trial k = 6;
- Probability of success on a single trial p = 0.9.
P(n=6; k=6; p=0.9) = 0.531441. ANSWER
b) none of them dislike their mother-in law?
- number of trials n = 6;
- number of success trial k = 0;
- Probability of success on a single trial p = 0.9.
P(n=6; k=0; p=0.9) = 0.000001. ANSWER
c) at least four of them dislike their mother-in law?
- number of trials n = 6;
- number of success trial k >= 4;
- Probability of success on a single trial p = 0.9.
P(n=6; k>=4; p=0.9) = 0.98415. ANSWER
d) no more than three of them dislike their mother-in law?
- number of trials n = 6;
- number of success trial k <= 3;
- Probability of success on a single trial p = 0.9.
P(n=6; k<=3; p=0.9) = 0.01585. ANSWER
Solved.