What is the Minimum Probability for Successful Ticket Purchases?

[asd1249jf]

Homework Statement

You are a manager of ticket agency that sells concert tickets. You assume that people will call 3 times in an attempt to buy the ticket and give up. You want to make sure that you are able to serve at least 95% of the people who wants tickets. Let p be the probability that a caller gets through to your ticket agency. What is the minimum value of p necessary to meet your goal?

Homework Equations

Theorem of Probability Mass Function

The Attempt at a Solution

So there are three independent outcome. Let X be the random variable for the corresponding three events, and let S denote the successful purchase of the ticket and F the failure case.

{X = 0} = {S} / Successful in first call
{X = 1} = {FS} / Successful in second call
{X = 2} = {FFS} / Successful in third call
{X = 3} = {FFF} / No Ticket purchase

I'm completely stuck here. Frankly, I don't even know if I'm approaching this problem correctly. Any help will be appriciated.

 

[gabbagabbahey]

If p is the probabality a caller gets through on each call,(1) what is the probability that the caller gets through the first time?

(2) what is the probability that the caller gets through the second time?

(3) what is the probability that the caller gets through the third time?

(4)what is the total probability?

Hint- for a caller to get through on the second call, he must have also failed to get through on the first call

 

[asd1249jf]

Ok. Obviously, the probability of getting the ticket on the first call is p.

Probability on the second one is 1-(1-p)^2 and the third one is 1-(1-p)^3

Then the total probability is p + (1-(1-p)^2) + (1-(1-p)^3)

Am I right? And from this, do I set the whole equation to 0.95 and solve for p? I don't quite understand the notion here

 

[gabbagabbahey]

You're right on the first one, but not the next two...what is the probability of failing the first time?

 

[asd1249jf]

Probability of failing the first time is 1-p

 

[gabbagabbahey]

Right, so the probability of getting through the second time is equal to the probability of failing the first time AND succeeding the second {FS} so wouldn't that just be (1-p)p?

In other words F=1-p , S=p. What does that make the probability that a caller gets through on the 3rd attempt?

 

[asd1249jf]

oh ok, then the next event is {FFS] so it's (1-p)^2 * p for the third attempt.

Ok, then the total probability is p + p(1-p) + p(1-p)^2

So I set this expression equal to 0.95 and solve for p?

 

[gabbagabbahey]

Yup you got it, you want the total probability of {X=0}+{X=1}+{X=2} (as you have them labelled) to be at least 0.95, so just solve {X=0}+{X=1}+{X=2}=0.95.

Alternatively, you know that {X=0}+{X=1}+{X=2}+{X=3}=1 since the total probability of all possible outcomes must be 1. So you could have saved a little time by noting that {X=0}+{X=1}+{X=2}=1-{X=3} and just found {X=3}={FFF} and solved the equation 1-{FFF}=0.95. Either way you will get the same thing.

 

[asd1249jf]

Ahhhh! I should've noticed that earlier! I was like "I have to solve a third-order equation in a probability class??"

haha thanks a lot.

 

[gabbagabbahey]

Ahhhh! I should've noticed that earlier! I was like "I have to solve a third-order equation in a probability class??"

haha thanks a lot.

No problem, but it is still a 3rd order equation you should get.