Question

Here's the given scenario:

The percentage of American men who say they would marry the same woman if they had to do it all over again is 80%. The percentage of American women who say they would marry the same man again is 50%. (Source: Harper's Index).

So, the question is:

1) In a group of 10 married men, what is the probability that at least 7 will claim that they would marry the same woman again? 2) What about the probability that less than 4 will say this? Explain your answer.

Answer 1

This can be solved using the binomial distribution. Are you familiar with the binomial distribution?

Answer 2

@arturomoises1 Are you there?

Answer 3

Sorry I had logged off! You have helped me tremendously. I have the formulas, I just needed to know which one to use...Binomial Distribution is accurate.

Answer 4

1)
\[P(7)=C(10, 7) \times0.8^{7}\times0.2^{3}\]
\[P(8)=C(10, 8) \times0.8^{8}\times0.2^{2}\]
\[P(9)=C(10, 9) \times0.8^{9}\times0.2\]
\[P(10)=0.8^{10}\]
P(at least 7)= P(7) + P(8) + P(9) + P(10) = you can calculate

Answer 5

2)
\[P(0)=0.2^{10}\]
\[P(1)=C(10, 1) \times0.8\times0.2^{9}\]
\[P(2)=C(10, 2) \times0.8^{2}\times0.2^{8}\]
\[P(3)=C(10, 3) \times0.8^{3}\times0.2^{7}\]
P(fewer than 4)=P(0) + P(1) + P(2) + P(3) = you can calculate