Thread: Geometric distribution

For some clarification and pointers thanks.

The telephone lines coming into a helpdesk are occupied 55% of the time

a. If you are calling this helpdesk, what is the probability that it will take you more than three calls to get through?

I believe this is 0.55^3?

b. If, in addition, a friend tries independently to contact the same helpdesk, what is the probability that it will require more than three calls between you?

Any pointers to help me get started here? thanks

Originally Posted by Geometor

For some clarification and pointers thanks.

The telephone lines coming into a helpdesk are occupied 55% of the time

a. If you are calling this helpdesk, what is the probability that it will take you more than three calls to get through?

I believe this is 0.55^3?

No. One of 4 things can happen. 1) You get through on the first call: probability 1- .55= .45. 2) you get through on the second call: meaning you did NOT get through on the first call, prob. .55, but did on the second, prob .45. The probability of that is (.55)(.45)= .2475. 3) you get through on the third call. By the same argument the probabilty of that is (.55)(.55)(.45)= .136125. 4) It takes more than four calls. The probability you got through on the first, second, or third calls is 0.45+ 0.2475+ 0.1356125= 0.833625. The probability you do NOT get through in three calls is 1 minus that: 1- 0.833625= 0.166375.

b. If, in addition, a friend tries independently to contact the same helpdesk, what is the probability that it will require more than three calls between you?

Exactly the same answer. It doesn't matter who is calling!

Any pointers to help me get started here? thanks

Yes, study! Memorize the basic formulas.