Question

probability and statistics
the spinner is equally likely to land on any of the five sections
what is the probability that the spinner lands on an even number or on the unshaded section?

Explanation:

Step1: Identify the total number of sections

The spinner has 5 sections. So, the total number of possible outcomes \( n(S)=5 \).

Step2: Identify the favorable outcomes

The even - numbered sections are 2 and 4. The un - shaded section is 3. But we need to use the formula for the probability of \( A\cup B\) (where \(A\) is the event of landing on an even number and \(B\) is the event of landing on the un - shaded section).
The formula for \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\).
For event \(A\) (even number): \(n(A) = 2\) (sections 2 and 4), so \(P(A)=\frac{2}{5}\).
For event \(B\) (un - shaded section): \(n(B)=1\) (section 3), so \(P(B)=\frac{1}{5}\).
Since \(A\cap B=\varnothing\) (an even number and the un - shaded section 3 have no common elements), \(P(A\cap B) = 0\).
Then \(P(A\cup B)=\frac{2 + 1-0}{5}=\frac{3}{5}\).

Answer:

\(\frac{3}{5}\)