# Conditional probability

#### mhet

##### New Member
Hi! I'm working on a binomial conditional probability question and I am stuck. I would really appreciate some help. The question is as follows: You have a box of n=40 bolts. The probability of a bolt being faulty is p=0.005. You are told that at least one of the bolts is faulty, and asked: What is the probability of more than one bolt being faulty given that at least one is faulty?

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.