Question

greg tossed a number cube and recorded the results. the table shows the number of times he rolled each number. based on the table, what is the experimental probability of rolling a 4 on the next toss?
gregs number cube tosses

number	frequency
2	5
3	6
4	4
5	8
6	6

(options: 1/9, 1/6, 4/9, 1/7)

Explanation:

Step1: Calculate total trials

Sum frequencies: \(7 + 5 + 6 + 4 + 8 + 6 = 36\)

Step2: Find frequency of 4

Frequency of 4 is \(4\)

Step3: Compute experimental probability

Experimental probability = \(\frac{\text{Frequency of 4}}{\text{Total trials}} = \frac{4}{36} = \frac{1}{9}\) Wait, no, wait. Wait the frequencies: Let's re - sum. 1:7, 2:5, 3:6, 4:4, 5:8, 6:6. So 7 + 5=12, 12 + 6 = 18, 18+4 = 22, 22 + 8=30, 30+6 = 36. Frequency of 4 is 4. So probability is 4/36=1/9? Wait but maybe I misread the table. Wait the table: Number 1:7, 2:5, 3:6, 4:4, 5:8, 6:6? Wait no, maybe the 5's frequency is 8? Wait the user's table: "Number 1:7, 2:5, 3:6, 4:4, 5:8, 6:6"? Wait no, looking at the image again, maybe the 6's frequency is 6? Wait the original table:

Greg’s Number Cube Tosses

Number | Frequency

1 | 7

2 | 5

3 | 6

4 | 4

5 | 8

6 | 6

Wait, sum is 7 + 5+6 + 4+8 + 6=36. So experimental probability of 4 is 4/36 = 1/9? But the options: Let's check the options. The options are \(\frac{1}{9}\), \(\frac{1}{8}\), \(\frac{1}{7}\), \(\frac{1}{6}\)? Wait no, maybe I misread the frequency of 4. Wait maybe the frequency of 4 is 6? Wait no, the table as per the user: "4 | 4", "5 | 8", "6 | 6". Wait maybe the user made a typo, but according to the given table, frequency of 4 is 4. So 4/36=1/9. So the answer is \(\frac{1}{9}\).

Answer:

\(\frac{1}{9}\)