Question

A spinner is divided into four equal sections labeled 1, 2, 3, and 4. Another spinner is divided into three equal sections labeled A, B, and C. Simon will spin each spinner one time. How many of the possible outcomes have an even number or a B?

Answer 1

Since we are asked to compute the probability of two event with "or"
statement,  then we will add the probabilities like this:
Denote A the first event and B the second event:
$P(A\cup B)=P(A)+P(B)$
A="Get an even number":
$P(A)= \frac{2}{4}= \frac{1}{2}$
B="Get section B":
$P(B)= \frac{1}{3}$
Conclusion:
$P(A\cup B)=P(A)+P(B)= \frac{1}{2} +\frac{1}{3}\\=\frac{5}{6}$
The probability is 5/6

Answer 2

Answer: 3/4

Step-by-step explanation:

The possible outcomes that are available in spinning each spinner once are A1, A2, A3, A4, B1, B2, B3, B4, C1, C2, C3, C4.

A2, A4, B1, B2, B3, B4, C2, C4 all have either an even number in their outcome, or a B. This is 8 of the 12 possible outcomes that follow the requirements listed. 8/12 is simplified to 3/4 of the outcomes that have either an even number or a B.