1. ## well driller

I am posting this problem here since I got to turn it in 2 hours:
It was originally posted in the probability forum for college.

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)

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

my calculator does not give me a approximate value for b(3.6;4,0.6) !
How should I treat this question?
Thank you for your time .
B

I am posting this problem here since I got to turn it in 2 hours:
It was originally posted in the probability forum for college.

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)
P(X>4) = P(6) + P(5) = b(6;6,0.6) + b(5;6,0.6) = 0.6^6 + 6 0.4 0.6^5 ~= 0.23328

2-P(X=4)
P(X=4)=b(4;6,0.6)
yes: b(4; 6, 0.6) = 6!/(4! 2!) ) 0.6^4 0.4^2 ~= 0.31104

3-mean
mean=n*p=3.6
yes.

4-variance
Var^2=n*p*(1-p)
Var = n p (1-p) = 6*0.6*0.4 ~= 1.44

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

my calculator does not give me a approximate value for b(3.6;4,0.6) !
How should I treat this question?
Since X cannot take non-integer values P(X=3.6) = 0.

RonL

3. Thanks captainBlack!