With n=40, p=0.005 and X=faulty bolts, I see this as: P(X=>2 | X=>1).

I figure that the two events are dependent as you cannot have more than two faulty bolts unless at least one is faulty.

To calculate this, I think I need to know:

A) The probability of no faulty bolts, P(X=0):

P(X=0) = (1-p)^n

P(X=0) =(1-0.005)^40 = 0.8183

B) The probability of at least one faulty bolt, P(X=>1):

Using the complementary rule, this is the same as 1 minus the probability of no faulty bolts:

P(X=>1)= 1 - (1-p)^n

P(X=>1) = 1 - 0.8183 = 0.1817

C) Also need to know the probability of exactly one bolt being faulty, P(X=1):

Following binomial probability calculation:

D) The probability of more than 1 faulty bolts, P(X=>2), is the same as 1 minus the probability of 1 or less faulty bolts, 1-P(X=<1):

And so P(X =>2) is:

And now I am unsure where to go from here. I think something like this is what I need to calculate:

but I can't figure out the correct way to calculate .

I would really appreciate some guidance, thanks.