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Binomial Distribution Statistics Problem

  • #1

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) = nCk * [tex]p^{k}[/tex] * [tex](1-p)^k{}[/tex]

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.
 

Answers and Replies

  • #2
76
0
Could you possibly tell us a little more about the variables in that formula. "What do C, k, n, and p represent?"
 
  • #3
623
0
Look up the birthday problem, it's pretty similar to this isn't it?
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618

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) = nCk * [tex]p^{k}[/tex] * [tex](1-p)^k{}[/tex]

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.
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,833
956

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) = nCk * [tex]p^{k}[/tex] * [tex](1-p)^k{}[/tex]

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) = nCk * [tex]p^{k}[/tex] * [tex](1-p)^{n-k}[/tex]


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, [itex](1- p)^{5-5}= (1-p)^0= 1[/itex] so it is just "probability of all the same" is [itex]p^5[/itex]. What is p?
 

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