Binomial Distribution Statistics Problem

[blondsk8rguy]

Homework Statement

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.)

Homework Equations

P(X=k) = _nC_k * $$p^{k}$$ * $$(1-p)^k{}$$

The Attempt at a Solution

I think that it involves in some way subtracting the probability that everyone has a unique zodiacal sign from 1, but I'm not sure exactly how.

 

[sonofjohn]

Could you possibly tell us a little more about the variables in that formula. "What do C, k, n, and p represent?"

 

[NoMoreExams]

Look up the birthday problem, it's pretty similar to this isn't it?

 

[Dick]

Homework Statement

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.)

Homework Equations

P(X=k) = _nC_k * $$p^{k}$$ * $$(1-p)^k{}$$

The Attempt at a Solution

I think that it involves in some way subtracting the probability that everyone has a unique zodiacal sign from 1, but I'm not sure exactly how.

Exactly the right approach. How many ways are there to assign a different sign to each person and how many ways to assign any sign to any person. Take the ratio and subtract from 1.

 

[HallsofIvy]

Homework Statement

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.)

Homework Equations

P(X=k) = _nC_k * $$p^{k}$$ * $$(1-p)^k{}$$

The Attempt at a Solution

I think that it involves in some way subtracting the probability that everyone has a unique zodiacal sign from 1, but I'm not sure exactly how.

Your formula is incorrect- perhaps a typo. It should be
P(X=k) = _nC_k * $$p^{k}$$ * $$(1-p)^{n-k}$$


Yes, "at least two the same" is the opposite of "all different". Since there are 5 people and you want 5 different signs, both n and k in your binomial coefficient are 5 so that is easy- its just 1. In fact, $(1- p)^{5-5}= (1-p)^0= 1$ so it is just "probability of all the same" is $p^5$. What is p?