Question

Another game on the “Price is Right” requires the contestant to spin a wheel with numbers 5, 10, 15, 20, .... , 100. What is the probability that the contestant spins 100 or 30?

Short answer

The probability is$\frac{1}{10}$

Step-by-step solution

Step 1. Given Information

The given data is that the contestant to spin a wheel with numbers 5, 10, 15, 20, ... , 100.

Step 2. Calculation  

The number of elements in the sample space are $n\left(S\right)=20$

The probability of event E "spins 100" is $P\left(E\right)=\frac{1}{20}$

The probability of event F "spins 30" is $P\left(F\right)=\frac{1}{20}$

Thus, the probability that the contestant spins 100 or 30 is as follows,

$$\begin{gathered} P(100or30)=P\left(E\right)+P\left(F\right) \\ =\frac{1}{20}+\frac{1}{20} \\ =\frac{2}{20} \\ =\frac{1}{10} \end{gathered}$$