Chapter 13.6, Problem 7CYU

To determine

The odd alphabet in favour of picking an $A,E,I,O,U$ .

Answer to Problem 7CYU

The odds in favour of getting an $A,E,I,O,U$ is $5:21$ .

Explanation of Solution

On alphabets there are $26$ letters present. So the odds in favour of getting an $A,E,I,O,U$ such as that is 5 outcomes is given as,

  $\begin{array}{l}P\left(\text{getting}\text{an}A,E,I,O,U\right)=\frac{\text{No}\text{.}\text{of}\text{ways}A,E,I,O,U\text{can}\text{occure}}{\text{No}\text{.}\text{of}\text{ways}A,E,I,O,U\text{cannot}\text{occure}}\\ P\left(\text{getting}\text{an}A,E,I,O,U\right)=\frac{A,E,I,O,U}{\text{Total}\text{No}\text{.}\text{of}\text{alphabets}\left(\text{getting}\text{an}A,E,I,O,U\right)}\\ P\left(\text{getting}\text{an}A,E,I,O,U\right)=\frac{5}{\left(26-5\right)}\\ P\left(\text{getting}\text{an}A,E,I,O,U\right)=\frac{5}{21}\approx 5:21\end{array}$

So, the odds in favour of getting $A,E,I,O,U$ is $5:21$ .

Conclusion:

Hence, the odds in favour of getting an $A,E,I,O,U$ is $5:21$ .