An oil drilling company ventures into various locations, and their success or failure is independent from one location to another. Suppose that the probability of a success at any specific location is .25. Also suppose that the drilling company feels that it will “hit it big” if the second success occurs before the sixth attempt. What is the probability that the driller will “hit it big.”

My solution:

P(success) = 0.25
P(2successes in 5trials) = C(5,2) * (0.25)^2 * (0.75)^3 = 0.2637

Does this make any sense? I think it's right.

Any input will be much appreciated.

Thanks

Could also get 3,4 or 5 in 5

making it

5c2 * 0.25^2 *0.75^3
5c3 * 0.25^3 *0.75^2
5c4 * 0.25^4 *0.75^1
5c5 * 0.25^5 *0.75^0

Add them all together should give your answer

Then do 1 and 0 by the same way and they should all add up to 1

Sorry haven't got my calc with me

Good Luck with your drilling!!

But the question states:
"Also suppose that the drilling company feels that it will “hit it big” if the second success occurs before the sixth attempt.What is the probability that the driller will “hit it big.”"

Does that not mean that the answer should only be for 2 success in 5 trials?

I think the question is a bit fuzzy. But I would go with your answer, ankitj. I think the drilling team would take some time off after they 'hit it big' - so there would be no need for 3, 4, or 5 successes. But seriously, I think a good problem would try to specifically imply 'greater than' if it were seeking more than 2 successes.

Yea, this professor is not very clear with his questions. Not the first time this has happenend. Thanks for your help though.

If they strike oil in the first 2 goes and then stop the probablility for that calculation is not

0.25^2 x 0.75^3 as the last 3 are not completed so you would have to use

First 2 hit = 0.25^2 = 0.0625

Then hit either 1 and 3 or 2 and 3 =

2 x 0.25^2 x 0.75 = 0.09375

Then (1,4 or 2,4 or 3,4) =

3 x 0.25^2 x 0.75^2 = 0.10546875

Then (1,5 or 2,5 or 3,5 or 4,5)

4 x 0.25^2 x 0.75^3 = 0.10546875

= 0.3671875

What do you think?

hmm that makes more sense. Thanks for your help.