Question

the following values are the results from a simulation for rolling a standard die 24 times. use the simulated results to estimate the probability of rolling either a 2 or a 3. how does this compare to the theoretical probability of rolling either of these numbers?
6 6 2 3 6 1 2 5 5 5 3 5 1 4 2 4 5 2 3 2 1 6 2 6
(1 point)
the simulated probability is 0.375, which is more than the theoretical probability.
the simulated probability is 0.533, which is more than the theoretical probability.
the simulated probability is 0.333, which is the same as the theoretical probability.
the simulated probability is 0.600, which is the same as the theoretical probability.

Explanation:

Step1: Count the number of 2s and 3s

Count the number of 2s and 3s in the simulation results.
Number of 2s: 7
Number of 3s: 3
Total number of 2s and 3s: \(7 + 3=10\)

Step2: Calculate the simulated probability

The formula for probability is \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\).
Here, the total number of outcomes \(n = 24\), and the number of favorable outcomes \(m=10\).
So the simulated probability \(P_{simulated}=\frac{10}{24}=\frac{5}{12}\approx0.417\) (Wait, let's re - check the count.
Looking at the sequence: 6 6 2 3 6 1 2 5 5 5 3 5 1 4 2 4 5 2 3 2 1 6 2 6.
Counting 2s: positions 3,7,15,18,19,23. So 6 2s. Counting 3s: positions 4,11,19 (but 19 is 2). Wait, no:
Let's list them out:
The sequence is: [6,6,2,3,6,1,2,5,5,5,3,5,1,4,2,4,5,2,3,2,1,6,2,6]
Number of 2s: 7 (positions 3,7,15,18,20,23,24? No, wait the sequence has 24 elements. Let's recount:

• Element 3: 2
• Element 4:3
• Element7:2
• Element11:3
• Element15:2
• Element18:2
• Element19:3
• Element20:2
• Element23:2
• Element24:6. So number of 2s:7 (elements 3,7,15,18,20,23) and number of 3s:3 (elements4,11,19). Total of 2s and 3s: \(7 + 3=10\)

Simulated probability \(P_{simulated}=\frac{10}{24}=\frac{5}{12}\approx0.417\) (Wrong approach. Wait, the formula is correct.
The theoretical probability:
A standard die has 6 faces. The probability of rolling a 2 is \(P(2)=\frac{1}{6}\), and the probability of rolling a 3 is \(P(3)=\frac{1}{6}\).
The theoretical probability of rolling a 2 or 3 is \(P_{theoretical}=P(2)+P(3)=\frac{1 + 1}{6}=\frac{1}{3}\approx0.333\)

Answer:

The simulated probability is \(P_{simulated}=\frac{10}{24}\approx0.417\) (Wait, no. Wait, \(\frac{10}{24}=\frac{5}{12}\approx0.417\) is wrong. Wait, \(10\div24 = 0.416\cdots\). But let's check the options.
If we calculate the number of 2s and 3s again:
The sequence: 6,6,2,3,6,1,2,5,5,5,3,5,1,4,2,4,5,2,3,2,1,6,2,6
Number of 2s: 9 (wait no, let's count one - by - one:

• First 2: position3
• Second 2: position7
• Third 2: position15
• Fourth 2: position18
• Fifth 2: position20
• Sixth 2: position23
• Seventh 2: no. Wait, no, the sequence:

1:6, 2:6, 3:2, 4:3, 5:6, 6:1, 7:2, 8:5,9:5,10:5,11:3,12:5,13:1,14:4,15:2,16:4,17:5,18:2,19:3,20:2,21:1,22:6,23:2,24:6
Number of 2s:3,7,15,18,20,23 → 6 2s
Number of 3s:4,11,19 → 3 3s. Total of 2s and 3s:9. So \(P_{simulated}=\frac{9}{24}=0.375\)
The theoretical probability \(P_{theoretical}=\frac{2}{6}=\frac{1}{3}\approx0.333\)
So the simulated probability is \(0.375\), which is more than the theoretical probability.
So the answer is: The simulated probability is \(0.375\), which is more than the theoretical probability. (The first option: The simulated probability is \(0.375\), which is more than the theoretical probability)