Question

Conditional Probability Question:

Components of a certain type are shipped to a supplier in batches of 10. Suppose that 50% of all such batches have no defective components, 30% have one defective component, and 20% have two defective components.

Two components from a batch are randomly selected. What are the probabilities associated with 0, 1 and 2 defective components being in the batch under each of the following conditions:

a) neither tested component is defective
b) One of the two components is defective

Answer 1

I'm unsure of where to start here. I know that the percentages are A, but I don't know what B is. I tried creating a tree but got lost.

Answer 2

I figured P(A1) = 0.5, P(A2) = 0.3, and P(A3) = 0.20. I don't know how to find B though. Is it supposed to be the number of defective components you're given taken out of 10?

Answer 3

Answers are:
a) neither test component defective:
0 defective components in batch = 0.578
1 defective components in batch = 0.278
2 defective components in batch = 0.144

b) one test component defective:
0 defective components in batch = 0
1 defective components in batch = 0.475
2 defective components in batch = 0.543

Answer 4

Right. And 0 components out of 10 being defective is just another way of saying all 10 components are in working order. P(A') = 1-0 = 1 or whatever.

Answer 5

Ugh...this book is bloody confusing when it comes to wording.

Answer 6

It's a strange question, since it seems to imply there will never be more than 2 defective components despite there being batches of 10.

Answer 7

Unless they only ever test two, and the 'two components' really means 'two or more components'; however, they ought to specify that if it were the case.

Answer 8

Note that the question does not state whether the sample of two is taken with or without replacement.

Answer 9

http://math.stackexchange.com/questions/415802/defective-components-binomial-distribution

Answer 10

Let \(D_i\) be the event there are \(i\) defective components in a batch.
They give us:\[
\Pr(D_0) = 0.5\\\Pr(D_1) = 0.3\\ \Pr(D_2) = 0.2
\]Let \(T_i\) be the event that \(i\) tested components are defective.
This would mean that (a) asks for \(T_0\) and (b) asks for \(T_1\).

The law of total probability tells us that: \[
T_i = \sum_{k=0}^2\Pr(T_i|D_k)\Pr(D_k)
\]We know \(T_i|D_k\) is the event that \(i\) tested components are defective if there are \(k\) defects.
It must be the case that \(i\leq k\), so the probability would be \(0\) otherwise.

If \(k=0\):\[
\Pr(T_0|D_0) = 1,\ \Pr(T_1|D_0)=0
\]If \(k=1\):\[
\Pr(T_0|D_1) = \frac{9}{10}\cdot \frac{8}{9},\ \Pr(T_1|D_1)=\frac{1}{10}\cdot \frac{9}{9}
\]If \(k=2\):\[
\Pr(T_0|D_2) = \frac{8}{10}\cdot \frac{7}{9},\ \Pr(T_1|D_2)=\frac{2}{10}\cdot \frac{8}{9}
\]Note that the general equation is given by: \[
\Pr(T_i|D_k) =\begin{cases}
\frac{(k\text{ P }i)((10-k)\text{ P }(2-i))}{10 \text{ P } 2} &i\leq k\\
0&\text{otherwise}
\end{cases}
\]Where \(n\text{ P }r\) denotes permutations: \(\frac{n!}{(n-r)!}\)

Answer 11

My answers are: \[
T_0=1\cdot \frac{5}{10} + \frac{9}{10}\cdot \frac{8}{9}\cdot \frac{3}{10}+\frac{8}{10}\cdot \frac{7}{9}\cdot \frac{2}{10} = \frac{389}{450} = 0.86\bar 4
\]and\[
T_1=\frac{1}{10}\cdot\frac{9}{9}\cdot \frac{3}{10} + \frac{2}{10}\cdot \frac{8}{9}\cdot \frac{2}{10} = \frac{59}{900} = 0.06\bar 5
\]

Answer 12

Your avatar is amusing.