Answer:
The reasonable prediction for successful rolls is 4.
Explanation:
Assuming the rolling cube is a fair 6 sided cube, so the probability of success of one roll is given as
![P(Success)=(Number\ of\ successful\ events)/(Number\ of\ total\ events)](https://img.qammunity.org/2022/formulas/mathematics/high-school/tms1qq2dhgvtrq5etiv8xizydc9ddaz5br.png)
![P(Success)=(1)/(6)](https://img.qammunity.org/2022/formulas/mathematics/high-school/g9ngl7ka7uk2zky5qgbnavhdfqci8dmqbf.png)
The total success is given as
![P(Total\ Success)=P(Success)* Number\ of\ events](https://img.qammunity.org/2022/formulas/mathematics/high-school/i1gkq4fzbeggfzr4xwum7vowo6i7p4w30u.png)
For 24 rolls it is given as
![P(Total\ Success)=P(Success)* Number\ of\ events\\P(Total\ Success)=(1)/(6)* 24\\P(Total\ Success)=4](https://img.qammunity.org/2022/formulas/mathematics/high-school/1fsw7m7stitwbp88slnrz3xhn9adudbxm5.png)
So the reasonable prediction for successful rolls is 4.