TITLE: Defective items probability question.
QUESTION [7 upvotes]: Hi I'm working with probability as part of an engineering course, and I'm struggling with the following tutorial question: 
Components of a certain type are shipped to a supplier in batches of ten. Suppose that 50% of all such batches contain no defective components, 35% contain one defective component while 15% contain two defective components. If two components are randomly selected from the batch. What are the probabilities associated with 0, 1 and 2 defective components being in the batch under each of the following conditions?
a) Neither selected component is defective.
b) One of the two components is defective.
c) Both components are defective.
I've considered using hypergeometric probability distribution for events P(35%) and P(15%) while comparing to P(50%) but this yields no result.

REPLY [3 votes]: Try using Bayes' theorem: $$P(A|X) = P(X|A) \dfrac{P(A)}{P(X)}$$
For instance, let event $A_m$ be "there are $m$ defective components in the batch" and let event $X_n$ be "$n$ of the selected components are defective".  Then we have the given marginal probabilities for $A_m$:
$$\begin{aligned}
P(A_0) &= 0.5 & P(A_1) &= 0.35 & P(A_2) &= 0.15
\end{aligned}$$
and the conditional probabilities for $X_n$ given $A_m$:
$$\begin{aligned}
P(X_0|A_0) &= 1 & P(X_0|A_1) &= \frac45 & P(X_0|A_2) &= \frac{28}{45} \\
P(X_1|A_0) &= 0 & P(X_1|A_1) &= \frac15 & P(X_1|A_2) &= \frac{16}{45} \\
P(X_2|A_0) &= 0 & P(X_2|A_1) &= 0 & P(X_2|A_2) &= \frac1{45} 
\end{aligned}$$
(Exercise: Why those probabilities? It's just a simple case of sampling without replacement.)
From the conditional probabilities, we can also calculate the marginal probabilities for $X_n$ using the law of total probability:
$$\begin{eqnarray}
P(X_0) &= 1\cdot0.5 &+& \frac45\cdot0.35 &+& \frac{28}{45}\cdot0.15 =& \frac{131}{150} =& 0.87333\dots \\
P(X_1) &= 0\cdot0.5 &+& \frac15\cdot0.35 &+& \frac{16}{45}\cdot0.15 =& \frac{37}{300} =& 0.12333\dots \\
P(X_2) &= 0\cdot0.5 &+& 0\cdot0.35 &+& \frac{1}{45}\cdot0.15 =& \frac{1}{300} =& 0.00333\dots \\
\end{eqnarray}$$
Now apply Bayes' theorem to get the conditional probabilities $P(A_m|X_n)$.