Question

On the “Price is Right” there is a game in which a bag is filled with 3 strike chips and 5 numbers. Let’s say that the numbers in the bag are 0, 1, 3, 6, and 9. What is the probability of selecting a strike chip or the number 1?

Short answer

The probability is$\frac{1}{2}$

Step-by-step solution

Step 1. Given Information

The given data is that the numbers in the bag are 0, 1, 3, 6, and 9.

Step 2. Calculation  

The number of elements in the sample space are $n\left(S\right)=3+5=8$

The event $E_1$consists of all the strike chips. So, probability of selecting a strike chip is $P\left(E_1\right)=\frac{3}{8}$

The event $E_2$consists of only the number 1. So, probability of selecting number 1 is role="math" localid="1647005899037" $P\left(E_2\right)=\frac{1}{8}$

Thus, the probability of selecting a strike chip or number 1 is as follows,

$$\begin{gathered} P\left(E\right)=P\left(E_1\right)+P\left(E_2\right) \\ =\frac{3}{8}+\frac{1}{8} \\ =\frac{4}{8} \\ =\frac{1}{2} \end{gathered}$$