Chapter 6.7, Problem 18E

To determine

To explain how to find the odds in favour of and the odds against the event.

Answer to Problem 18E

If the probability of the event is $P\left(A\right)$ then odd in favour is $\frac{P\left(A\right)}{1-P\left(A\right)}$ and odd against is $\frac{1-P\left(A\right)}{P\left(A\right)}$ .

Explanation of Solution

Given information :

The probability of an event is $P\left(A\right)=\frac{\text{Number of favourable outcome}}{\text{Number of possible outcome}}$ .

Let the probability of the event be $P\left(A\right)$ . Also,

  $\begin{array}{l}P\left(A\right)=\frac{\text{Number of favourable outcome}}{\text{Number of possible outcome}}\\ P\left(A\right)=\frac{\text{Number of favourable outcome}}{\text{Number of favourable outcome}+\text{Number of unfavourable outcome}}\\ \frac{1}{P\left(A\right)}=\frac{\text{Number of favourable outcome}+\text{Number of unfavourable outcome}}{\text{Number of favourable outcome}}\\ \frac{1}{P\left(A\right)}=\frac{\text{Number of favourable outcome}}{\text{Number of favourable outcome}}+\frac{\text{Number of unfavourable outcome}}{\text{Number of favourable outcome}}\end{array}$

  $\begin{array}{l}\frac{1}{P\left(A\right)}=\text{1}+\frac{\text{Number of unfavourable outcome}}{\text{Number of favourable outcome}}\\ \frac{1}{P\left(A\right)}=\text{1}+\text{Odd against the event}\\ \frac{1}{P\left(A\right)}-1=\text{Odd against the event}\\ \frac{1-P\left(A\right)}{P\left(A\right)}=\text{Odd against the event}\end{array}$

  $\text{Odd against the event}=\frac{1-P\left(A\right)}{P\left(A\right)}$

Therefore, odd against the event A is $\frac{1-P\left(A\right)}{P\left(A\right)}$ .

And odd in favour of vent A will be $\frac{P\left(A\right)}{1-P\left(A\right)}$ .

Hence,

If the probability of the event is $P\left(A\right)$ then odd in favour is $\frac{P\left(A\right)}{1-P\left(A\right)}$ and odd against is $\frac{1-P\left(A\right)}{P\left(A\right)}$ .