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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 seven married women are having coffee together one morning. Find the following probabilities. (For each answer, enter a number. 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.

Sagot :

For each married women, there are only two possible outcomes. Either they dislike their mother in law, or they do not. The probability of a women disliking their mother in law is independent of any other women, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

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).

This means that [tex]p = 0.9[/tex]

Suppose that seven married women are having coffee together one morning.

This means that [tex]n = 7[/tex]

Question a:

This is P(X = 7), so:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 7) = C_{7,7}.(0.9)^{7}.(0.1)^{0} = 0.4783[/tex]

Thus, 0.4783 = 47.83% probability that all of them dislike their mother-in-law.

Question b:

This is P(X = 0), so:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{7,0}.(0.9)^{0}.(0.1)^{7} = 10^{-7}[/tex]

Thus, [tex]10^{-7}[/tex] probability that none of them dislike their mother-in-law.

Question c:

This is:

[tex]P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7)[/tex], so:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 5) = C_{7,5}.(0.9)^{5}.(0.1)^{2} = 0.1240[/tex]

[tex]P(X = 6) = C_{7,6}.(0.9)^{6}.(0.1)^{1} = 0.3720[/tex]

[tex]P(X = 7) = C_{7,7}.(0.9)^{7}.(0.1)^{0} = 0.4783[/tex]

Then

[tex]P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) = 0.1240 + 0.3720 + 0.4783 = 0.9743[/tex]

0.9743 = 97.43% probability that at least five of them dislike their mother-in-law.

Question d:

This is:

[tex]P(X \leq 4) = 1 - P(X > 4)[/tex]

Considering [tex]P(X > 4) = P(X \geq 5) = 0.9743[/tex]:

[tex]P(X \leq 4) = 1 - P(X > 4) = 1 - 0.9743 = 0.0257[/tex]

Thus, 0.0257 = 2.57% probability that no more than four of them dislike their mother-in-law.

For another example of the binomial distribution, you can check https://brainly.com/question/15557838

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