Geometric Probability Distribution - (Y > 10)

Problem:

An oil prospector drills a succession of holes to find a productive well. The probability he is successful on a given trial is .2. If the prospector can afford to drill at most ten wells, what is the probability that he will fail to find a productive well?

Here's what I have so far:

This is a geometric prob. dist. because the sample space contains the countably infinite set of sample points.

p = 0.2

P(Y > 10)

= 1 - [ P(Y = 1), P(Y = 2), ... , P(Y = 10) ]

I know there is an easier way to do this than adding up the first 10 events, but that's where I'm stuck. I found a similar problem in my textbook where P(Y >= 3), and it had the formula:

1- p - qp.

If I do that for this problem: 1 - (0.2) - (0.8)(0.2) = 0.64. I know that value is way too high.

Does anybody know what I'm doing wrong? Thanks.

Re: Geometric Probability Distribution - (Y > 10)

Quote:

Originally Posted by

**crossingdouble** Problem:

An oil prospector drills a succession of holes to find a productive well. The probability he is successful on a given trial is .2. If the prospector can afford to drill at most ten wells,

what is the probability that** he will **__fail__ to find a productive well?

$\displaystyle (0.8)^{10}$

Re: Geometric Probability Distribution - (Y > 10)

Thanks Plato, I got it now.

I found the correct formula:

P(Y >= n) = 1 - (1 - ((1 - p)^n))

For this problem, it simplifies to (0.8)^10.

And, your approach is actually more simple.