Find the geometric probability of an event with a probability of 0.2 having its first success on the fourth trial.

Question

Find the geometric probability of an event with a probability of 0.2 having its first success on the fourth trial.

Answer 1

On any individual trial P(success) = 0.2

Therefore P(failure) = 1 - P(success) = 0.8  

 

In order to have the first success on the fourth trial there must be 3 failures followed by 1 success.  

 

The probability of this sequence occurring is (0.8) * (0.8) * (0.8) * (0.2) = 0.1024

 

Note that this is (1-P(success))^3 * P(success) = (0.8)^3 * (0.2) 

 

 

Answer 2

OK, I'm not sure what "geometric probability" means in this situation, but normally, this would be done as follows:

 

In order to have its first success on its fourth trial, it has to fail on the first three.  The probability of failure is .8.

 

So the probability of failing on the first three and then succeeding on the fourth is .8 * .8 * .8 * .2 = .1024