Geometric Probabilities

with Practice Problems

Often we face situations where we repeat trials until we achieve success, then stop.

For example, Larry will ask a succession of girls to go with him to the high school prom.

Once a girl says yes, he stops asking.


The outcome looks something like this:


F     F     F     F     S


where F is a failure and S a success.

So here, Larry asks 5 girls, and the fifth one agrees to go with him.

This is a binomial situation with n = 5 trials and X = 1 success. With one exception.



There are \( _5C_1\) = 5 possible ways to get 1 success from 5 trials:

S     F     F     F     F

F     S     F     F     F

F     F     S     F     F

F     F     F     S     F

F     F     F     F     S

But only the last one applies here.

S     F     F     F     F

F     S     F     F     F

F     F     S     F     F

F     F     F     S     F

F     F     F     F     S

So we omit Part A of the binomial formula, because there is only one way to get your one success on the last trial.

That leaves just Part B.


prob(X) = \(p^X \times (1-p)^{n-X}\)


And because X always equals 1 when you stop after the first success, the probability becomes:


prob(X = 1) = \(p \times (1-p)^{n-1}\)


This is a geometric probability. It is a special case of the binomial where

  • X = 1, and
  • the number of ways to get X = 1 success, after n - 1 failures, is 1.


So we characterize the probability of getting that one, lone success on the nth trial as:

prob(n) = \(p \times (1-p)^{n-1}\)


All that is needed are n and p.


If the probability of Larry getting a girl to say yes in one trial is, for example, 1 in 10, or 0.10, then the probability of getting a yes on the 5th trial (n = 5) after getting 4 noes (yep, that's the plural of no) is: 


prob(5) = \(p \times (1-p)^{n-1}\)

                  = \(0.10 \times (0.90)^{4}\)     

= \(0.0656\)


Click the right arrow to try a couple of practice problems.

Problem 1


Jessica wants to see the mysterious Marfa lights. For each trip to Marfa, the chances of her seeing the lights is only about 1 in 12, or 0.0833.


She's already made 7 trips. What is the probability that she'll see them this next trip if she's not seen them on the previous 7 trips?


(Click the down arrow to see the solution.)

Problem 1 Solution


This would be Jessica's 8th trip, so n = 8. So we have:

        prob(n = 8) = \(p \times (1-p)^{n-1}\)

                            = \(0.0833 \times (0.9167)^7\) 

                            = \(0.0453\)


(Click the right arrow for Problem 2.)

Problem 2

Kurt is trying to win a big stuffed animal for his wife. For one dollar, he can try and knock over a small pyramid of bottles with a baseball.

The probability of succeeding with one ball (one trial) is one out of fifty, or 0.02. Kurt has already spent 29 dollars. He wants to try one more time.

What's the probability he can hit it on the 30th attempt after missing 29 times?


(Click the down arrow.)

Problem 2 Solution



        prob(n = 30) = \(p \times (1-p)^{n-1}\)

                              = \(0.02 \times (0.98)^{29}\) 

                              = \(0.0111\)


The Geometric Distribution

By smilinjoe

The Geometric Distribution

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