Chapter 5, Problem 12PT

(a)

To determine

To draw: Tree diagram for the given data.

Explanation of Solution

Given:

Percentage of items produced by machine A $=60\%=0.60$

Percentage of items produced by machine B $=30\%=0.30$

Percentage of items produced by machine C $=10\%=0.10$

Percentage of items produced by machine A is defective $=10\%=0.10$

Percentage of items produced by machine B is defective $=30\%=0.30$

Percentage of items produced by machine C is defective $=40\%=0.40$

Formula used:

The Bayes theorem states that

  $P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}$

Graph:

Let A denote the event of an item produced by machine A.

B denote the event of an item produced by machine B.

C denote the event of an item produced by machine C.

D denote the event of an item being defective.

ND be the event of an item being not defective

The tree diagram with the associated probabilities is as follows

  

(b)

To determine

To find: The probability of a part being defective.

Answer to Problem 12PT

The probability is 0.19.

Explanation of Solution

Calculation:

As per the sub part a

  $\begin{array}{l}P\left(A\right)=0.60\\ P\left(B\right)=0.30\\ P\left(C\right)=0.10\end{array}$

The conditional probabilities are

  $\begin{array}{l}P\left(D|A\right)=0.10\\ P\left(D|B\right)=0.30\\ P\left(D|C\right)=0.40\end{array}$

Using the above formula

  $\begin{array}{l}P\left(D\cap A\right)=0.60\times 0.10=0.06\\ P\left(D\cap B\right)=0.30\times 0.30=0.09\\ P\left(D\cap C\right)=0.10\times 0.40=0.04\end{array}$

Therefore,

  $\begin{array}{l}P\left(D\right)=0.06+0.09+0.04\\ P\left(D\right)=0.19\end{array}$

Thus, the required probability is 0.19.

(c)

To determine

To find: The machine which is most likely to have produced the defective item given a defective item was chosen.

Answer to Problem 12PT

Machine B

Explanation of Solution

Formula used:

Calculation:

From sub part b, $P\left(D\right)=0.19$

The probability that machine A has produced the defective item which is chosen

  $\begin{array}{l}P\left(A|D\right)=\frac{P\left(A\cap D\right)}{P\left(D\right)}\\ P\left(A|D\right)=\frac{0.06}{0.19}=0.3158\end{array}$

The probability that machine B has produced the defective item which is chosen

  $\begin{array}{l}P\left(B|D\right)=\frac{P\left(B\cap D\right)}{P\left(D\right)}\\ P\left(B|D\right)=\frac{0.09}{0.19}=0.4737\end{array}$

The probability that machine C has produced the defective item which is chosen

  $\begin{array}{l}P\left(C|D\right)=\frac{P\left(C\cap D\right)}{P\left(D\right)}\\ P\left(AC|D\right)=\frac{0.04}{0.19}=0.2105\end{array}$

Conclusion:

As machine B has the highest probability so, machine is most likely to have produced the defective item which is chosen.