Question

in a single experiment, a die is tossed and a spinner with the letters a, b, and c is spun. each letter is equally likely. a. the sample space is 9. the probability of getting a 2 or a b. the probability of getting a b is $\frac{1}{9}$. b. the sample space is 18. the probability of getting a 2 or a b is $\frac{18}{8}=\frac{9}{4}$. c. the sample space is 18. the probability of getting a 2 or a b is $\frac{8}{18}=\frac{4}{9}$. d. the sample space is 9. the probability of getting a 2 or a b is $\frac{8}{18}=\frac{4}{9}$.

Explanation:

Step1: Calculate sample - space size

A die has 6 outcomes and a spinner with 3 letters has 3 outcomes. By the fundamental counting principle, the sample - space size $n(S)=6\times3 = 18$.

Step2: Define events

Let $A$ be the event of getting a 2 on the die and $B$ be the event of getting a B on the spinner.
$n(A)=3$ (since for a 2 on the die, there are 3 possible spinner outcomes: (2,A), (2,B), (2,C)). $n(B)=6$ (since for each of the 6 die - outcomes, there is a possibility of getting B on the spinner: (1,B), (2,B), (3,B), (4,B), (5,B), (6,B)). And $n(A\cap B) = 1$ (the outcome (2,B)).

Step3: Use the addition rule of probability

The formula for $P(A\cup B)$ is $P(A\cup B)=\frac{n(A)+n(B)-n(A\cap B)}{n(S)}$.
$n(A)+n(B)-n(A\cap B)=3 + 6-1=8$.
$P(A\cup B)=\frac{8}{18}=\frac{4}{9}$.

Answer:

C. The sample space is 18. The probability of getting a 2 or a B is $\frac{8}{18}=\frac{4}{9}$.