Chapter 12.8, Problem 2P

You spin a spinner that has 12 equalized sections numbered 1 to 12. Find each probability.

P(even or 7)

To determine

Find the probability :

P(even or 7)

Answer to Problem 2P

  $\frac{7}{12}$

Explanation of Solution

Given:

You spin a spinner that has 12 equal-sized sections numbered 1 to 12.

Formula Used:

  $P(E)=\frac{\text{Favourable outcomes of E}}{\text{total number of outcomes}}$

P(A or B) = P(A) + P(B) , if A and B are mutually exhaustive events, that is if both don’t have anything in common.

Calculation:

Total number of sections = 12

Number of favorable outcomes that are even numbers = 6 ={2,4,6,8,10,12}

Number of favorable outcomes for 7 = 1

  $P(even)=\frac{\text{favourable outcomes }}{total\text{ number of outcomes}}=\frac{6}{12}=\frac{1}{2}$

  $P(7)=\frac{\text{favourable outcomes for 7}}{total\text{ number of outcomes}}=\frac{1}{12}$

Since getting an even number and getting a 7 are mutually exhaustive events, so,

  $P(evenor\text{ 7})=P(even)+P(7)=\frac{1}{2}+\frac{1}{12}=\frac{7}{12}$