What is the probability of getting a 1 when a fair die is tossed? A fair die is tossed 500 times - Junior Cycle Mathematics - Question a - 2012

The number of times a 3 appeared can be calculated as follows:

Total tosses = 500

The total frequency of the other numbers is:

$70 + 82 + 90 + 91 + 81 = 414$

Hence, the number of times a 3 appeared is:

$500 - 414 = 86$

So, the answer is 86.

To calculate the relative frequency for each outcome, we can use the formula:

$\text{Relative Frequency} = \frac{\text{Frequency}}{\text{Total Frequency}}$

Calculating for each outcome:

1. For Outcome 1: $\frac{70}{500} = 0.14$
2. For Outcome 2: $\frac{82}{500} = 0.16$
3. For Outcome 3: $\frac{86}{500} = 0.17$
4. For Outcome 4: $\frac{90}{500} = 0.18$
5. For Outcome 5: $\frac{91}{500} = 0.18$
6. For Outcome 6: $\frac{81}{500} = 0.16$

Therefore, the relative frequencies rounded to two decimal places are:

• Outcome 1: 0.14
• Outcome 2: 0.16
• Outcome 3: 0.17
• Outcome 4: 0.18
• Outcome 5: 0.18
• Outcome 6: 0.16

The difference in value between the relative frequency for 1 in the table and the calculated probability from part (a) could be attributed to experimental error. Since this is based on a limited number of tosses (500), random variations can affect the frequency of outcomes. Additionally, the law of large numbers states that as the number of trials increases, the experimental probability will converge towards the theoretical probability, which is $\frac{1}{6}$ for a fair die.