Question

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? 1/5 2/5 3/5 4/5

Explanation:

Step1: Define events

Let $A$ be the event that the spinner lands on an even - number. Let $B$ be the event that the spinner lands on the unshaded section. The sample space $n(S)=5$.

Step2: Find $n(A)$ and $P(A)$

The even - numbered sections are 2 and 4. So $n(A) = 2$, and $P(A)=\frac{n(A)}{n(S)}=\frac{2}{5}$.

Step3: Find $n(B)$ and $P(B)$

Assume the unshaded section is one of the 5 sections. So $n(B)=1$, and $P(B)=\frac{n(B)}{n(S)}=\frac{1}{5}$.

Step4: Find $n(A\cap B)$ and $P(A\cap B)$

If the unshaded section is not an even - numbered section, $n(A\cap B) = 0$, and $P(A\cap B)=0$.

Step5: Use the addition rule of probability

The addition rule of probability is $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Substitute $P(A)=\frac{2}{5}$, $P(B)=\frac{1}{5}$, and $P(A\cap B) = 0$ into the formula: $P(A\cup B)=\frac{2 + 1-0}{5}=\frac{3}{5}$.

Answer:

$\frac{3}{5}$