Question

the spinner below is spun once. find each probability as a fraction in simplest form.

1. ( p(\text{shaded}) )
2. ( p(\text{unshaded or even}) )
3. ( p(\text{prime and less than 10}) )

© gina wilson (all things algebra®, llc), 2020

Explanation:

Step1: Determine total sections

Count the number of equal - sized sections on the spinner. By looking at the numbers (1,2,3,4,5,6,7,8,9,10,11,12), we can see that there are 12 sections in total.

Step2: Solve for \(P(\text{shaded})\)

First, count the number of shaded sections. Let's assume from the spinner (by looking at the shaded and unshaded parts) that the number of shaded sections is 8.
The formula for probability is \(P(E)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\).
So, \(P(\text{shaded})=\frac{8}{12}=\frac{2}{3}\) (simplify by dividing numerator and denominator by 4).

Step3: Solve for \(P(\text{unshaded or even})\)

• First, find the number of unshaded sections: Total sections = 12, shaded = 8, so unshaded = \(12 - 8=4\) (sections with numbers 1,5,6,9? Wait, let's list the numbers: Unshaded sections: Let's assume the unshaded regions have numbers 1,5,6,9? Wait, no, let's list all numbers and their status (shaded or unshaded) and even/odd.

Wait, let's list the numbers on the spinner: 1 (unshaded), 2 (shaded), 3 (shaded), 4 (shaded), 5 (unshaded), 6 (unshaded), 7 (shaded), 8 (shaded), 9 (unshaded), 10 (shaded), 11 (shaded), 12 (shaded).
Now, unshaded numbers: 1,5,6,9 (4 numbers). Even numbers: 2,4,6,8,10,12 (6 numbers). Now, the formula for \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), where \(A\) is unshaded and \(B\) is even.

• \(P(A)=\frac{4}{12}\), \(P(B)=\frac{6}{12}\), \(A\cap B\) (unshaded and even) is the number 6, so \(P(A\cap B)=\frac{1}{12}\).
• Then \(P(\text{unshaded or even})=\frac{4}{12}+\frac{6}{12}-\frac{1}{12}=\frac{4 + 6-1}{12}=\frac{9}{12}=\frac{3}{4}\).

Step4: Solve for \(P(\text{prime and less than 10})\)

• Prime numbers less than 10: Prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves. The numbers on the spinner less than 10 are 1,2,3,4,5,6,7,8,9.
• Prime numbers in this set: 2,3,5,7. Let's check the spinner: 2 (shaded), 3 (shaded), 5 (unshaded), 7 (shaded). So there are 4 prime numbers less than 10.
• Using the probability formula \(P(E)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\), so \(P(\text{prime and less than 10})=\frac{4}{12}=\frac{1}{3}\).

Answer:

1. \(\frac{2}{3}\)
2. \(\frac{3}{4}\)
3. \(\frac{1}{3}\)