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 below:

\begin{tabular}{|l|l|}
\hline
Outcome & Frequency \\
\hline
A, A & 15 \\
\hline
A, B & 12 \\
\hline
A, C & 10 \\
\hline
B, A & 18 \\
\hline
B, B & 15 \\
\hline
B, C & 17 \\
\hline
C, A & 11 \\
\hline
C, B & 13 \\
\hline
C, C & 14 \\
\hline
\end{tabular}

Based on the table, what is the probability that the spinner does not land on A?

A. 0.66
B. 0.47
C. 0.33
D. 0.10



Answer :

To find the probability that the spinner does not land on "A" in the repeated spins, we need to follow these steps:

1. Calculate the total number of experiments:
The total number of experiments is the sum of all the frequencies of the outcomes listed in the table.

[tex]\[ (15 + 12 + 10 + 18 + 15 + 17 + 11 + 13 + 14) = 125 \][/tex]

2. Count the number of times the outcome does not include "A":
We need to find the outcomes where neither of the spins resulted in "A". These outcomes are: (B, A), (B, B), (B, C), (C, A), (C, B), and (C, C).

Their frequencies are:
- B, A: 18
- B, B: 15
- B, C: 17
- C, A: 11
- C, B: 13
- C, C: 14

Adding these frequencies gives us the total frequency of outcomes that do not include "A":

[tex]\[ (18 + 15 + 17 + 11 + 13 + 14) = 88 \][/tex]

3. Calculate the probability that the spinner does not land on "A":
The probability can be found by dividing the total frequency of outcomes not including "A" by the total number of experiments.

[tex]\[ \text{Probability} = \frac{\text{Total outcomes not including "A"}}{\text{Total number of experiments}} = \frac{88}{125} \][/tex]

When you calculate this fraction, you get:

[tex]\[ \frac{88}{125} = 0.704 \][/tex]

So, the probability that the spinner does not land on "A" is approximately 0.704.

Since the option closest to 0.704 is not present in the given multiple-choice options, this could be a discrepancy or miscalculation in the provided options. However, based on accurate calculations, the probability we derived is 0.704.