Answer:
The probability of getting an odd number when the spinner is spun, is p(even number) = 1/2
Step-by-step explanation:
Given that the spinner has 12 equal sections labelled 1-12
Let S be the sample space
Then
S = {1,2,3,4,5,6,7,8,9,10,11,12}
[tex]n(S) = 12[/tex]
Let A be the event that the outcome is an even number
Then
A = {2,4,6,8,10,12}
[tex]n(A) = 6[/tex]
The probability of A will be given as:
[tex]p(A) = \frac{n(A)}{n(S)}\\\\= \frac{6}{12}\\\\= \frac{1}{2}[/tex]
Hence,
The probability of getting an odd number when the spinner is spun, is p(even number) = 1/2
