# Math Help - Probability

1. ## Probability

Past experience has shown that, on average, only 1 in 10 wells drilled hits oil. Assume the events are independent.

What is the probability that out of 5 drills, the company has 2 that hit oil?

What is the probability that the second hit was on the fifth drill.

Some help would be appreciated, thanks.

2. Originally Posted by Len
Past experience has shown that, on average, only 1 in 10 wells drilled hits oil. Assume the events are independent.

What is the probability that out of 5 drills, the company has 2 that hit oil?
The probability of a hit is $\frac{1}{10}$ and a miss is $\frac{9}{10}$, so if we want two hits, then we get $\left( \frac{1}{10} \right)^2 \left( \frac{9}{10} \right)^3 =$

Originally Posted by Len
What is the probability that the second hit was on the fifth drill.

Some help would be appreciated, thanks.

This one is a Negative Binomial Probability Distribution, which is defined as:

$\binom{y-1}{r-1}p^r(1-p)^{y-r}$ therefore we have:

y=5, r=2, $p=\tfrac{1}{10}$

then:

$\binom{5-1}{2-1}\left(\frac{1}{10}\right)^2\left( 1- \frac{1}{10} \right)^{5-2}=$

3. Originally Posted by Len
Past experience has shown that, on average, only 1 in 10 wells drilled hits oil. Assume the events are independent.

What is the probability that out of 5 drills, the company has 2 that hit oil?

[snip]
Originally Posted by lllll
The probability of a hit is $\frac{1}{10}$ and a miss is $\frac{9}{10}$, so if we want two hits, then we get $\left( \frac{1}{10} \right)^2 \left( \frac{9}{10} \right)^3 =$

[snip]
Small correction: It will be ${\color{red}{5 \choose 2}} \left( \frac{1}{10} \right)^2 \left( \frac{9}{10} \right)^3 \, ....$

4. Originally Posted by Len
Past experience has shown that, on average, only 1 in 10 wells drilled hits oil. Assume the events are independent.

[snip]

What is the probability that the second hit was on the fifth drill.

Some help would be appreciated, thanks.
This one can also be done as follows:

Pr(exactly 1 hit in first four drills) x Pr (hit on fifth drill), where:

Pr(exactly 1 hit in first four drills) = ${4 \choose 1} \left(\frac{1}{10}\right)^1 \left(\frac{9}{10}\right)^3$.

Pr (hit on fifth drill) = 1/10.