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.

Fin