1. ## Binomial Distribution Problems

I believe that all of these are binomial distribution problems, but I'm not really sure how to approach them. Any help would be appreciated greatly.

1. Estimate the probability that, in a group of five people, at least two of them have the same zodiacal sign. (There are 12 zodiacal signs; assume that each sign is equally likely for any person.)

2. 30% of the workers in a workforce are women. A company hires 100 workers of which 25 are women. What is the probability this (the hiring of 24 women or less) occurred by chance?

3. To ensure a high male/female ratio, the ruler of a mythical island decrees couples may keep having children until they have a girl. If the decree is followed, what will the male/female ratio be on the island?

2. Originally Posted by blondsk8rguy
I believe that all of these are binomial distribution problems, but I'm not really sure how to approach them. Any help would be appreciated greatly.

1. Estimate the probability that, in a group of five people, at least two of them have the same zodiacal sign. (There are 12 zodiacal signs; assume that each sign is equally likely for any person.)

Mr F says: First assume a particular zodiac sign (Aries, say) and calculate the probability that at least two in the group have that sign. Binomial where n = 5, p = 1/12. Now multiply that answer by 12 (why?).

2. 30% of the workers in a workforce are women. A company hires 100 workers of which 25 are women. What is the probability this (the hiring of 24 women or less) occurred by chance?

Mr F says: Binomial where n = 100, p = 0.3. Calculate ${\color{red}\Pr(X \leq 25)}$. You could probably use the normal approximation if you don't have access to technology.

3. To ensure a high male/female ratio, the ruler of a mythical island decrees couples may keep having children until they have a girl. If the decree is followed, what will the male/female ratio be on the island?
3. Presumambly there is no more than 1 girl per family. So you need to calculate the expected number of boys in a family. So use the geoemtric distribution (Geometric distribution - Wikipedia, the free encyclopedia - use a support of k = 0, 1, 2, 3, ....) to get the expected number of births until a girl is obtained. The answer may surprise you.

3. Thanks for your help. I didn't realize that my teacher had said that the problems could use geometric distributions. Here's my progress on each of the questions:

1. I used a binomcdf function with n=5 and p=1/12 ranging from 2 to 5 to account for combinations of 2, 3, 4, or all 5 people sharing the same sign. The multiplication by 12 is due to the possibility of any zodiac combination, and comes from nCr(12, 1), right? So I got about 70%, which seems somewhat high, but is still reasonable.

2. This problem seems really straightforward now.... I just performed a binomcdf function for n=100 and p=.3 from 0 to 25. This gave about 16%, which seems low but is probably reasonable.

3. The answer to this problem was really surprising.... When I found the expected value of the geometric distribution, I got 1. Does that mean that the ratio ends up being 1:1?

There's also a second part to the first question, which involves estimating the probability that at least one of the five people has the same zodiacal sign as yours. For this, would I do a binomcdf operation for n=5 and p=1/12 from 1 to 5 and not multiply by 12? That gives about 35%, which seems to make sense.

Thanks so much.

4. Originally Posted by blondsk8rguy
Thanks for your help. I didn't realize that my teacher had said that the problems could use geometric distributions. Here's my progress on each of the questions:

1. I used a binomcdf function with n=5 and p=1/12 ranging from 2 to 5 to account for combinations of 2, 3, 4, or all 5 people sharing the same sign. The multiplication by 12 is due to the possibility of any zodiac combination, and comes from nCr(12, 1), right? So I got about 70%, which seems somewhat high, but is still reasonable.

2. This problem seems really straightforward now.... I just performed a binomcdf function for n=100 and p=.3 from 0 to 25. This gave about 16%, which seems low but is probably reasonable.

3. The answer to this problem was really surprising.... When I found the expected value of the geometric distribution, I got 1. Does that mean that the ratio ends up being 1:1?

There's also a second part to the first question, which involves estimating the probability that at least one of the five people has the same zodiacal sign as yours. For this, would I do a binomcdf operation for n=5 and p=1/12 from 1 to 5 and not multiply by 12? That gives about 35%, which seems to make sense.

Thanks so much.
Everything you say is correct. (For Q3 it's the expected ratio)

5. Originally Posted by blondsk8rguy
I believe that all of these are binomial distribution problems, but I'm not really sure how to approach them. Any help would be appreciated greatly.

1. Estimate the probability that, in a group of five people, at least two of them have the same zodiacal sign. (There are 12 zodiacal signs; assume that each sign is equally likely for any person.)

Mr F says: First assume a particular zodiac sign (Aries, say) and calculate the probability that at least two in the group have that sign. Binomial where n = 5, p = 1/12. Now multiply that answer by 12 (why?).

2. 30% of the workers in a workforce are women. A company hires 100 workers of which 25 are women. What is the probability this (the hiring of 24 women or less) occurred by chance?

Mr F says: Binomial where n = 100, p = 0.3. Calculate ${\color{red}\Pr(X \leq 25)}$. You could probably use the normal approximation if you don't have access to technology.

3. To ensure a high male/female ratio, the ruler of a mythical island decrees couples may keep having children until they have a girl. If the decree is followed, what will the male/female ratio be on the island?
3. Presumambly there is no more than 1 girl per family. So you need to calculate the expected number of boys in a family. So use the geoemtric distribution (Geometric distribution - Wikipedia, the free encyclopedia - use a support of k = 0, 1, 2, 3, ....) to get the expected number of births until a girl is obtained. The answer may surprise you.
My answer to Q1 is very wrong. I don't have time now but will post a correct solution later.

6. Originally Posted by blondsk8rguy
I believe that all of these are binomial distribution problems, but I'm not really sure how to approach them. Any help would be appreciated greatly.

1. Estimate the probability that, in a group of five people, at least two of them have the same zodiacal sign. (There are 12 zodiacal signs; assume that each sign is equally likely for any person.)
[snip]
Pr(at least two people in a group of five have the same zodiac sign) = 1 - Pr(no people in a group of five have the same zodiac sign).

Pr(no people in a group of five have the same zodiac sign) $= 1 \cdot \left(1 - \frac{1}{12} \right) \cdot \left(1 - \frac{2}{12} \right) \cdot \left(1 - \frac{3}{12} \right) \cdot \left(1 - \frac{4}{12} \right)$

$= \left(\frac{11}{12} \right) \cdot \left(\frac{10}{12} \right) \cdot \left(\frac{9}{12} \right) \cdot \left(\frac{8}{12} \right) = \, ....$

It is left as a simple exercise to explain why my original answer is wrong.