# Panel problem

Hi,
I need have this problem check because I have a problem at the last question:

the probability is 0.6 that a well driller will find a well of water at a depth less than 100 feet in a certain area. Wells are to be drilled for six new homeowners. Assue that finding a well of water at a depth of less than 100 feet is independent from drilling to drilling and that the probability is 0.6 on every drilling. If X is the number of wells of water found. Find:

1-P(X>4)
My solution: P(X>4)=1-B(4;6,0.6)

2-P(X=4)
P(X=4)=b(4;6,0.6)

3-mean
mean=n*p=3.6

4-variance
Var^2=n*p*(1-p)=1.44

5-P(X=mean)
Here since mean =3.6, I have problem to find P(X=mean)

Thank you
B.

## Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
Hi,
I need have this problem check because I have a problem at the last question:

the probability is 0.6 that a well driller will find a well of water at a depth less than 100 feet in a certain area. Wells are to be drilled for six new homeowners. Assue that finding a well of water at a depth of less than 100 feet is independent from drilling to drilling and that the probability is 0.6 on every drilling. If X is the number of wells of water found. Find:

1-P(X>4)
My solution: P(X>4)=1-B(4;6,0.6)
B is the binomial probability? Isn't B(4;6,0.6) the probability of exactly 4 "sucesses" out of 6? If so 1- B(4;6,0,6) is P(X= 1, 2, 3, 5, or 6). What you want is P(5;6,0.6)+ P(6;6,0.6).

2-P(X=4)
P(X=4)=b(4;6,0.6)
?? Is "b" different from "B"? If not then what I thought above was true. This is correct, 1 is incorrect. Oh, and have you actually calculated that value?

3-mean
mean=n*p=3.6

4-variance
Var^2=n*p*(1-p)=1.44
Okay.

5-P(X=mean)
Here since mean =3.6, I have problem to find P(X=mean)

Thank you
B.
Well, that last one is kind of trivial isn't it!

But my calculator does not give me a approximate value for b(3.6;4,0.6) ! ( for question 5)

HallsofIvy
Science Advisor
Homework Helper
You are supposed to be smarter than your calculuator! Stop and THINK. If you drill 6 wells, each well either will or won't hit water, what is the probability that exactly 3.6 of them will hit water?!!

You are supposed to be smarter than your calculuator! Stop and THINK. If you drill 6 wells, each well either will or won't hit water, what is the probability that exactly 3.6 of them will hit water?!!

Each well has a probability of 0.5 to hit water. So the probability of exactly 3.6 wells each 0.5^3.6= 0.082 ??