Experimental Probability Worksheet | Fun and Engaging PDF Worksheets - Free Printable

Here are the solutions to the problems on the worksheet.

Section A

Table Calculations:
For a standard fair die, the theoretical probability of rolling a '6' is always $\frac{1}{6}$ (approx 0.17).
*Note: Since you have to roll the dice yourself for this section, your "Number of '6's" and "Experimental probability" will be different from mine. Below is an example of how to fill it out.*

| Number of Trials | Number of '6's (Example) | Experimental Probability (Calculation) | Theoretical Probability |
| :--- | :--- | :--- | :--- |
| 10 | 2 | $2 \div 10 = \mathbf{0.20}$ | 0.17 |
| 20 | 3 | $3 \div 20 = \mathbf{0.15}$ | 0.17 |
| 30 | 6 | $6 \div 30 = \mathbf{0.20}$ | 0.17 |
| 40 | 5 | $5 \div 40 = \mathbf{0.13}$ | 0.17 |
| 50 | 9 | $9 \div 50 = \mathbf{0.18}$ | 0.17 |
| 60 | 11 | $11 \div 60 = \mathbf{0.18}$ | 0.17 |

Questions:

1) Were the results from your experiment as you expected? Why? Why not?
*Answer:* Results vary. Usually, with only 60 rolls, the experimental probability won't be exactly 0.17. It might be higher or lower because 60 is a small number of trials. Random chance causes variation in small experiments.

2) Which is the most reliable estimate of the experimental probability of scoring a '6'?
*Answer:* The result from 60 trials. As the number of trials increases, the experimental probability becomes more reliable and tends to get closer to the true theoretical probability.

3) How could you get an experimental probability that is closer to the theoretical probability of scoring a '6'?
*Answer:* You would need to increase the number of trials. Rolling the dice hundreds or thousands of times would make the experimental result much closer to 0.17.

***

Section B

Table Calculations:
Formula: $\text{Experimental Probability} = \frac{\text{Number of '6's}}{\text{Number of Rolls}}$

| Number of Rolls | Number of '6's | Experimental Probability | Theoretical Probability |
| :--- | :--- | :--- | :--- |
| 10 | 2 | $2 \div 10 = \mathbf{0.20}$ | 0.17 |
| 20 | 16 | $16 \div 20 = \mathbf{0.80}$ | 0.17 |
| 30 | 27 | $27 \div 30 = \mathbf{0.90}$ | 0.17 |
| 40 | 30 | $30 \div 40 = \mathbf{0.75}$ | 0.17 |
| 50 | 42 | $42 \div 50 = \mathbf{0.84}$ | 0.17 |
| 60 | 50 | $50 \div 60 = \mathbf{0.83}$ | 0.17 |

*(Note: Theoretical probability is kept constant at 0.17 for comparison).*

Questions:

1) What does the horizontal line (in red) represent?
*Answer:* It represents the theoretical probability of rolling a '6' on a fair dice, which is $\frac{1}{6}$ or approximately 0.17.

2) Is the dice a fair dice? How can you tell from your graph?
*Answer:* No, the dice is not fair. We can tell because the experimental probability (the plotted points) is very high (around 0.80 to 0.90) and stays far away from the theoretical line (0.17). A fair dice would have results hovering around 0.17. This dice seems to be weighted to land on '6' much more often.

3) How would the graph compare with the horizontal line (theory line) as the number of trials increase?
*Answer:* If this were a fair dice, the graph would move closer to the horizontal line as trials increased. However, since this dice is biased (unfair), the graph will stay far above the horizontal line. Even with more trials, the experimental probability will likely stay around 0.83 because the dice itself is rigged.

Final Answer:
See the completed tables and answers above.