Confusing conditional probability question I have a probability question, that's been bugging me for a while now:   

We are given that defective items on display occur independently with probability 0.05. A random sample of 100 items is taken. Given that at least 99 of the sampled items are not defective, I want to find the probability that the first item sampled is not defective.  

Please, how do I go about solving this question?  
Thanks. 
 A: Let $A$ be the event that the first item is not defective. Let $B$ be the event that at least 99 items are not defective. Then we want $$P~(A|B)=\frac{P(A~\text{and}~B)}{P~(B)}.$$
$$
\begin{align*} 
P~(B)& = P~(99~\text{are not defective}) + P~ (100 ~\text{are not defective})\\
&=100\cdot 0.95^{99}\cdot 0.05 + 0.95^{100}.

\end{align*}$$
$$\begin{align*}
P~(A~\text{and}~ B) &= P~(\text{one is defective and the first is not defective})\\
& +~ P~(\text{none is defective and the first is not defective})\\
& = 99\cdot 0.95^{99}\cdot 0.05 + 0.95^{100}.
\end{align*}$$
I hope this helps.
A: This can be solved by a straightforward application of Bayes' Theorem.  If 
$FD$ represents the event that the first sampled item is defective, and $ND$ represents the event that at least ninety-nine of the sampled items are not defective, then we get:
$$P(FD|ND)=\frac{P(ND|FD)P(FD)}{P(ND)}$$
With a little thought, you should be able to calculate the conditional probability and the two unconditional probabilities on the right hand side, given the information you have.  This looks like it probably needs a (homework) tag, so I'll leave it at that.
