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Sagot :
Let's break down the problem and find the probabilities step-by-step.
Theoretical Probability:
Since a fair die has six faces, the theoretical probability for each face (1 through 6) is 1/6.
- Theoretical Probability of rolling a 1 ( [tex]\( P(1) \)[/tex] ) = [tex]\( \frac{1}{6} \)[/tex] ≈ 0.1667
- Theoretical Probability of rolling a 2 ( [tex]\( P(2) \)[/tex] ) = [tex]\( \frac{1}{6} \)[/tex] ≈ 0.1667
- Theoretical Probability of rolling a 3 ( [tex]\( P(3) \)[/tex] ) = [tex]\( \frac{1}{6} \)[/tex] ≈ 0.1667
- Theoretical Probability of rolling a 4 ( [tex]\( P(4) \)[/tex] ) = [tex]\( \frac{1}{6} \)[/tex] ≈ 0.1667
- Theoretical Probability of rolling a 5 ( [tex]\( P(5) \)[/tex] ) = [tex]\( \frac{1}{6} \)[/tex] ≈ 0.1667
- Theoretical Probability of rolling a 6 ( [tex]\( P(6) \)[/tex] ) = [tex]\( \frac{1}{6} \)[/tex] ≈ 0.1667
Experimental Probability:
Given the frequency of each result from rolling the die 40 times, we calculate the experimental probability by dividing the frequency of each result by the total number of rolls (40).
- Experimental Probability of rolling a 1 ( [tex]\( P(1) \)[/tex] ) = [tex]\( \frac{2}{40} \)[/tex] = 0.05
- Experimental Probability of rolling a 2 ( [tex]\( P(2) \)[/tex] ) = [tex]\( \frac{10}{40} \)[/tex] = 0.25
- Experimental Probability of rolling a 3 ( [tex]\( P(3) \)[/tex] ) = [tex]\( \frac{5}{40} \)[/tex] = 0.125
- Experimental Probability of rolling a 4 ( [tex]\( P(4) \)[/tex] ) = [tex]\( \frac{4}{40} \)[/tex] = 0.10
- Experimental Probability of rolling a 5 ( [tex]\( P(5) \)[/tex] ) = [tex]\( \frac{7}{40} \)[/tex] = 0.175
- Experimental Probability of rolling a 6 ( [tex]\( P(6) \)[/tex] ) = [tex]\( \frac{12}{40} \)[/tex] = 0.30
Putting these into the table:
[tex]\[ \begin{array}{|c|c|c|c|} \hline \text{Result} & \text{Frequency} & \text{Theoretical Probability} & \text{Experimental Probability} \\ \hline 1 & 2 & 0.1667 & 0.05 \\ \hline 2 & 10 & 0.1667 & 0.25 \\ \hline 3 & 5 & 0.1667 & 0.125 \\ \hline 4 & 4 & 0.1667 & 0.10 \\ \hline 5 & 7 & 0.1667 & 0.175 \\ \hline 6 & 12 & 0.1667 & 0.30 \\ \hline \end{array} \][/tex]
This completes the solution for the given question showing both the theoretical and experimental probabilities for each result of a die rolled 40 times.
Theoretical Probability:
Since a fair die has six faces, the theoretical probability for each face (1 through 6) is 1/6.
- Theoretical Probability of rolling a 1 ( [tex]\( P(1) \)[/tex] ) = [tex]\( \frac{1}{6} \)[/tex] ≈ 0.1667
- Theoretical Probability of rolling a 2 ( [tex]\( P(2) \)[/tex] ) = [tex]\( \frac{1}{6} \)[/tex] ≈ 0.1667
- Theoretical Probability of rolling a 3 ( [tex]\( P(3) \)[/tex] ) = [tex]\( \frac{1}{6} \)[/tex] ≈ 0.1667
- Theoretical Probability of rolling a 4 ( [tex]\( P(4) \)[/tex] ) = [tex]\( \frac{1}{6} \)[/tex] ≈ 0.1667
- Theoretical Probability of rolling a 5 ( [tex]\( P(5) \)[/tex] ) = [tex]\( \frac{1}{6} \)[/tex] ≈ 0.1667
- Theoretical Probability of rolling a 6 ( [tex]\( P(6) \)[/tex] ) = [tex]\( \frac{1}{6} \)[/tex] ≈ 0.1667
Experimental Probability:
Given the frequency of each result from rolling the die 40 times, we calculate the experimental probability by dividing the frequency of each result by the total number of rolls (40).
- Experimental Probability of rolling a 1 ( [tex]\( P(1) \)[/tex] ) = [tex]\( \frac{2}{40} \)[/tex] = 0.05
- Experimental Probability of rolling a 2 ( [tex]\( P(2) \)[/tex] ) = [tex]\( \frac{10}{40} \)[/tex] = 0.25
- Experimental Probability of rolling a 3 ( [tex]\( P(3) \)[/tex] ) = [tex]\( \frac{5}{40} \)[/tex] = 0.125
- Experimental Probability of rolling a 4 ( [tex]\( P(4) \)[/tex] ) = [tex]\( \frac{4}{40} \)[/tex] = 0.10
- Experimental Probability of rolling a 5 ( [tex]\( P(5) \)[/tex] ) = [tex]\( \frac{7}{40} \)[/tex] = 0.175
- Experimental Probability of rolling a 6 ( [tex]\( P(6) \)[/tex] ) = [tex]\( \frac{12}{40} \)[/tex] = 0.30
Putting these into the table:
[tex]\[ \begin{array}{|c|c|c|c|} \hline \text{Result} & \text{Frequency} & \text{Theoretical Probability} & \text{Experimental Probability} \\ \hline 1 & 2 & 0.1667 & 0.05 \\ \hline 2 & 10 & 0.1667 & 0.25 \\ \hline 3 & 5 & 0.1667 & 0.125 \\ \hline 4 & 4 & 0.1667 & 0.10 \\ \hline 5 & 7 & 0.1667 & 0.175 \\ \hline 6 & 12 & 0.1667 & 0.30 \\ \hline \end{array} \][/tex]
This completes the solution for the given question showing both the theoretical and experimental probabilities for each result of a die rolled 40 times.
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