there is a mistake in the question .I am sorry they are asking for two flaws ,not 1.

- What is the expected number of panels that need to be sampled before two flaws are found.

I guess the steps are the same but for E(x)=2.

In this case is it still a geometric distribution?

Now for problem 2:

**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)

Thank you for your time