Answer:
![p(Not\ 5) = \frac{8}{9}](https://tex.z-dn.net/?f=p%28Not%5C%205%29%20%3D%20%5Cfrac%7B8%7D%7B9%7D)
Step-by-step explanation:
Given:
Txo six-sided dice are rolled.
Total number of outcomes n(S) = 36
We need to find the probability that the sum is not equal to 5 p(Not 5).
Solution:
Using probability formula.
----------------(1)
Where:
n(E) is the number of outcomes favourable to E.
n(S) is the total number of equally likely outcomes.
The sum of two six-sided dice roll outcome is equal to 5 as.
Outcome as 5: {(1,4), (2,3), (3,2), (4,1)}
So, the total favourable events n(E) = 4
Now, we substitute n(E) and n(s) in equation 1.
![P(5)=\frac{4}{36}](https://tex.z-dn.net/?f=P%285%29%3D%5Cfrac%7B4%7D%7B36%7D)
![p(5) = \frac{1}{9}](https://tex.z-dn.net/?f=p%285%29%20%3D%20%5Cfrac%7B1%7D%7B9%7D)
Using formula.
![p(Not\ E) + p(E) = 1](https://tex.z-dn.net/?f=p%28Not%5C%20E%29%20%2B%20p%28E%29%20%3D%201)
![p(Not\ 5) + p(5) = 1](https://tex.z-dn.net/?f=p%28Not%5C%205%29%20%2B%20p%285%29%20%3D%201)
Now we substitute p(5) in above equation.
![p(Not\ 5) + \frac{1}{9} = 1](https://tex.z-dn.net/?f=p%28Not%5C%205%29%20%2B%20%5Cfrac%7B1%7D%7B9%7D%20%3D%201)
![p(Not\ 5) = 1-\frac{1}{9}](https://tex.z-dn.net/?f=p%28Not%5C%205%29%20%3D%201-%5Cfrac%7B1%7D%7B9%7D)
![p(Not\ 5) = \frac{9-1}{9}](https://tex.z-dn.net/?f=p%28Not%5C%205%29%20%3D%20%5Cfrac%7B9-1%7D%7B9%7D)
![p(Not\ 5) = \frac{8}{9}](https://tex.z-dn.net/?f=p%28Not%5C%205%29%20%3D%20%5Cfrac%7B8%7D%7B9%7D)
Therefore, the sum of two six-sided dice roll outcome is not equal to 5.
![p(Not\ 5) = \frac{8}{9}](https://tex.z-dn.net/?f=p%28Not%5C%205%29%20%3D%20%5Cfrac%7B8%7D%7B9%7D)