Defective items probability question.

Question

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.

Answer 1

Let $B_0$ be the event that the batch has $0$ defectives, $B_1$ be the event the batch has $1$ defective, and $B_2$ be the event the batch has $2$ defectives.

Let $D_0$ be the event that neither selected component is defective. Problem (a) asks us to find the conditional probabilities $\Pr(B_0|D_0)$, $\Pr(B_1|D_0)$, and $\Pr(B_2|D_0)$. Now the hardest part, identifying precisely what we are after, has been done!

For the calculation, we use the general conditional probability formula $$\Pr(X|Y)\Pr(Y)=\Pr(X\cap Y).$$ Put $X=B_0$ and $Y=D_0$. We need $\Pr(D_0)$ and $\Pr(B_0\cap D_0)$.

The event $D_0$ can happen in three different ways: (i) Our batch of $10$ is perfect, and we get no defectives in our sample of two; (ii) Our batch of $10$ has $1$ defective, but our sample of two misses them; (iii) Our batch has $2$ defective, but our sample misses them. If it helps, draw a tree that shows the three different paths through which we can end up with no defectives.

For (i), the probability is $(0.5)(1)$. For (ii), the probability that our batch has $1$ defective is $0.35$. Given that it has $1$ defective, the probability that our sample misses it is $\binom{9}{2}/\binom{10}{2}$, which is $8/10$. So the probability of (ii) is $(0.35)(8/10)$. For (iii), the probability our batch has $2$ defective is $0.15$. Given that it has $2$ defective, the probability that our sample misses both is $\binom{8}{2}/\binom{10}{2}$, which is $56/90$. So the probability of (iii) is $(0.15)(56/90)$. We have therefore found that $$\Pr(D_0)=(0.5)(1)+(0.35)(8/10)+(0.15)(56/90).$$ The probability $\Pr(B_0\cap D_0)$ has been calculated during our calculation of $\Pr(D_0)$. It is $(0.5)(1)$. We conclude that $$\Pr(B_0|D_0)=\frac{(0.5)(1)}{(0.5)(1)+(0.35)(8/10)+(0.15)(56/90)}.$$ The rest of the calculations for $D_0$ are easy, we have all the information needed. We get $$\Pr(B_1|D_0)=\frac{(0.35)(8/10)}{(0.5)(1)+(0.35)(8/10)+(0.15)(56/90)}$$ and $$\Pr(B_2|D_0)=\frac{(0.15)(56/90)}{(0.5)(1)+(0.35)(8/10)+(0.15)(56/90)}.$$

Alternately, we could have used Bayes' Formula directly . I wanted to do it in the above more basic way so that the logic would be clear.

Now unfortunately we have to deal with (b) and (c). But (c) is trivial! For (b), the calculation is as above, a little simpler, because if our sample of two has a defective, it cannot come from a perfect batch.

Answer 2

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 $P(A_m|X_n)$.