# Binomial distribution

1. Apr 2, 2009

### brendan

Hi Guys,
I have been given the probability that a drill strikes oil in a region = 0.2.

I know that if I wanted to find the probabilty of say striking oil 3 times out of 5 wells

It would be 5Choose3 = 5!/((2!)(3!)) * (1/5)3* (4/5)2 = 0.0512

My question is how would I go about finding the probability that the third strike occurs on the fifth well?

regards
Brendan

2. Apr 2, 2009

### HallsofIvy

That would be the probability that two strikes were found in four wells times the probability that the fifth well is a strike.

3. Apr 2, 2009

### brendan

So,
It would be 4Choose2 * Px=0.2 = 4!/((2!)(2!)) * (1/5)2* (4/5)2 *(1/5) = 0.03072
regards
Brendan

4. Apr 6, 2009

### brendan

Is the Expected value of striking oil 3 successive times just 0.2 * 0.2 * 0.2

= 0.008 ?
regards
Brendan

5. Apr 6, 2009

### CRGreathouse

That's the probability of striking oil three successive times out of three tries. If you have five tries, the chance is higher:
SSSnn
SSSnS
SSSSn
SSSSS
nSSSn
nSSSS
nnSSS

where n is nothing and S is a strike. Your probability covers the first four cases, but the following three cases are also possible.

6. Apr 6, 2009

### brendan

So,
Let say I have 5 tries there are 5P3 = 20 ways of getting 3 in a row.
and there are 5! = 120 combinations. so would that make the probability 5P3 divide 5! = 1/6 ?

regards
Brendan

7. Apr 7, 2009

### brendan

Sorry guys,
I should have said how many wells would have to be drilled to strike oil 3 times in in succession.

I should be using the Negative binomial for that calculation.

It is the number of successes k divided by the probabilty of success p

so k/p = 3/0.2 = 15 wells drilled before success