Solving Probability Problem: Finding a Well of Water at a Depth of <100ft

[brad sue]

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.

 

[HallsofIvy]

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!

 

[brad sue]

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

 

[HallsofIvy]

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?!

 

[brad sue]

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 ??