Answer:
0.4
Explanation:
These are the relative frequencies of each face (data are missing in the text of the problem):
Number Showing on Top Face Frequency
1 0
2 3
3 3
4 6
5 3
6 5
The probability to obtain a certain number when throwing the dice is given by
![p(x_i)=(f_i)/(\sum f)](https://img.qammunity.org/2021/formulas/mathematics/high-school/rswhdunz7hszktdzx6mtakprn7coq9euzc.png)
where
is the relative frequency of the number
to occur
is the sum of the relative frequencies
Here the sum of the frequencies is:
![\sum f=0+3+3+6+3+5=20](https://img.qammunity.org/2021/formulas/mathematics/high-school/ktk9ecm01u62aynjyh6cm4z2f9e0ogwllj.png)
For number 5 here, we have:
(from the table)
So the probability of getting a 5 is
![p(5)=(3)/(20)](https://img.qammunity.org/2021/formulas/mathematics/high-school/p55j028q6ed5rwivq6tp7q1qa8a3ykp04u.png)
For number 6 here, we have
![f_6=5](https://img.qammunity.org/2021/formulas/mathematics/high-school/5p3cpmxpnrb0jg1tikn9uf55bbjtm3t0e5.png)
So the probability of getting a 6 is
![p(6)=(5)/(20)](https://img.qammunity.org/2021/formulas/mathematics/high-school/eeb4jm404n4ryqy2gbwra13k1e5bsrc7jg.png)
So the probability to obtain either a 5 or a 6 in the next rolling is:
![p(5\cup 6)=p(5)+p(6)=(3)/(20)+(5)/(20)=(8)/(20)=0.4](https://img.qammunity.org/2021/formulas/mathematics/high-school/lq3h2yyieypl6pzy3vwa1353igyvv9ji8q.png)