Expert: Sherry Wallin - 4/29/2011

Question
The probability of four people out of ten having a blue ticket at a charity ball is 0.2508. What is the probability,as a percent, of a random person having a blue ticket?

Answer
Hi Andrea~

This is a binomial probability distribution problem, specifically a Bernoulli Trial.

Recall P(x) = n!/[(n-x)!x!] *p^x*q^(n-x)  ***Keep in mind that q = 1-p

Here n is 10, and x is 4, so n-x = 10 - 4 = 6

We know that the probability of four out of ten people having a blue ticket is .2508, thus

P(x) = n!/[(n-x)!x!] *p^x*q^(n-x) =>P(x) = n!/[(n-x)!x!] *p^x*(1-p)^(n-x)

P(4) = 10!/(6!4!) *p^4*(1-p)^6 = .2508

What you need to do is simplify and and solve for p:  210*p^4*(1-p)^6 = .2508 =>

p^4*(1-p)^6 ~= .001194285714    where I use ~= to mean approximately equal

You need to solve for p. I would use a calculator. You could use Pascal's Triangle and get p^4-6p^5+15p^6-20p^7+15p^8-6p^9+p^10

so you would have to solve for p p^4-6p^5+15p^6-20p^7+15p^8-6p^9+p^10 - .001194285714 = 0

and end up using software or a calculator anyway. It ends up that p is approximately .4

You can check it in your problem like this:

P(4) = 10!/(6!4!) *.4^4*.6^6 ?= .2508

210*.4^4*.6^6 ?= .250822656 ~= .2508  it checks!


Math Prof


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