Question

A spinner is divided into three equal parts A, B, and C. The repeated experiment of spinning the spinner twice is simulated 125 times. A table of outcomes is shown.

Outcome Frequency
A, A 15
A, B 12
A, C 10
B, A 18
B, B 15
B, C 17
C, A 11
C, B 13
C, C 14

Based on the table, for what probability can you expect the spinner to not land on B?
0.10
0.33
0.40
0.66

Answer 1

To find the probability that the spinner does not land on B when spun twice, we first need to identify the outcomes from the table that do not include B.

The outcomes from the table are as follows:
- A, A
- A, B
- A, C
- B, A
- B, B
- B, C
- C, A
- C, B
- C, C

The outcomes that do not land on B are:
- A, A
- A, C
- C, A
- C, C

Now we can calculate the total frequency of the outcomes that do not include B:

- Frequency of A, A = 15
- Frequency of A, C = 10
- Frequency of C, A = 11
- Frequency of C, C = 14

Adding these frequencies gives:
\[
15 + 10 + 11 + 14 = 50
\]

The total number of spins conducted is 125. Thus, the probability that the spinner does not land on B is given by the ratio of the favorable outcomes (those not including B) to the total outcomes:

\[
P(\text{not B}) = \frac{\text{Number of outcomes not including B}}{\text{Total outcomes}} = \frac{50}{125} = 0.40
\]

Therefore, the probability that the spinner does not land on B is **0.40**.