the spinner is equally likely to land on any of the six sections. what is the probability that the spinner…

the spinner is equally likely to land on any of the six sections. what is the probability that the spinner will land on a number greater than 4 or on a shaded section? $\frac{2}{9}$ $\frac{1}{2}$ $\frac{2}{3}$ $\frac{5}{6}$

the spinner is equally likely to land on any of the six sections. what is the probability that the spinner will land on a number greater than 4 or on a shaded section? $\frac{2}{9}$ $\frac{1}{2}$ $\frac{2}{3}$ $\frac{5}{6}$

Answer

Explanation:

Step1: Define events

Let $A$ be the event of landing on a number greater than 4. $A={5,6}$, so $n(A) = 2$. Let $B$ be the event of landing on a shaded - section. $B={1,3,5}$, so $n(B)=3$. And $A\cap B = {5}$, so $n(A\cap B)=1$. The total number of sections $n(S)=6$.

Step2: Use the addition - rule of probability

The formula for $P(A\cup B)$ is $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Since $P(A)=\frac{n(A)}{n(S)}$, $P(B)=\frac{n(B)}{n(S)}$ and $P(A\cap B)=\frac{n(A\cap B)}{n(S)}$, we have $P(A)=\frac{2}{6}$, $P(B)=\frac{3}{6}$ and $P(A\cap B)=\frac{1}{6}$.

Step3: Calculate the probability

$P(A\cup B)=\frac{2}{6}+\frac{3}{6}-\frac{1}{6}=\frac{2 + 3-1}{6}=\frac{4}{6}=\frac{2}{3}$.

Answer:

$\frac{2}{3}$