Answer:
![P (even,\ then\ not\ 2) = (5)/(12)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/8ildh9q93bpr77fyrrdy39cnntsnjkjzgo.png)
Explanation:
When launching a numerical cube there are 6 possible results
S = {1, 2, 3, 4, 5, 6}
Among these possible outcomes only 3 are even numbers
S = {2, 4, 6}
So the probability of obtaining an even number is
![P = (3)/(6)\\\\P = (1)/(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ar3787hiv0niy7z40bwev8b4pxkamtjcyn.png)
Then, the probability of obtaining a number other than 2 is:
.
If 2 launches are made then the events are independent,
Thus:
P(even, then not 2) is
![P (even,\ then\ not\ 2) = (1)/(2) * (5)/(6)\\\\P (even,\ then\ not\ 2) = (5)/(12)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/5h3nwnj1tnbzrqunyqq80voxldkqk5t6wy.png)