Chapter 13.6, Problem 18PPS

To determine

The oddsin favour of randomly selecting part having a fault.

Answer to Problem 18PPS

The odds in favour of randomly selecting part having a fault is $1:14$ .

Explanation of Solution

Given:

Total parts: $60$

Randomly selected: $56$

In the data there are a total of $60$ parts and where $56$ are randomly have no faults and the rest 4 have faults. So, in the odds in favour of picking a part that have fault is given as,

  $\begin{array}{l}P\left(\text{getting}\text{an}\text{part that has no faults}\right)=\frac{\text{No}\text{.}\text{of}\text{ways}\text{where}\text{a}\text{part having faults}\text{occur}}{\text{No}\text{.}\text{of}\text{ways}\text{where}\text{a}\text{part does not having faults}\text{occur}}\\ P\left(\text{getting}\text{an}\text{part that has no faults}\right)=\frac{4}{\text{Total}\text{No}\text{.}\text{of parts}-\left(\text{no}\text{.}\text{of}\text{parts that have faults}\right)}\\ P\left(\text{getting}\text{an}\text{part that has no faults}\right)=\frac{4}{\left(60-4\right)}\\ P\left(\text{getting}\text{an}\text{part that has no faults}\right)=\frac{4}{56}\approx 1:14\end{array}$

So, the odds in favour of randomly selecting part having a fault is $1:14$ .

Conclusion:

Hence, the odds in favour of randomly selecting part having a fault is $1:14$ .