Answer:
The probability that he will roll a 2 or a 3
![P((E_{1} UE_{2} ) = \frac{1}{3}](https://tex.z-dn.net/?f=P%28%28E_%7B1%7D%20UE_%7B2%7D%20%29%20%3D%20%5Cfrac%7B1%7D%7B3%7D)
Step-by-step explanation:
<u><em>Step(i);-</em></u>
Dimas is playing a game that uses a number cube.
The sides of the cube are labelled from 1 to 6
The total number of Exhaustive cases
n(S) = {1,2,3,4,5,6} = 6
<u><em>Step(ii):-</em></u>
Let E₁ be the event of roll a '2'
n(E₁) = 1
The probability that Dimas wants to roll a '2'
![P(E_{1} ) = \frac{n(E_{1} )}{n(S)} = \frac{1}{6}](https://tex.z-dn.net/?f=P%28E_%7B1%7D%20%29%20%3D%20%5Cfrac%7Bn%28E_%7B1%7D%20%29%7D%7Bn%28S%29%7D%20%3D%20%5Cfrac%7B1%7D%7B6%7D)
Let E₂ be the event of roll a '3'
n(E₂) = 1
The probability that Dimas wants to roll a '3'
![P(E_{2} ) = \frac{n(E_{2} )}{n(S)} = \frac{1}{6}](https://tex.z-dn.net/?f=P%28E_%7B2%7D%20%29%20%3D%20%5Cfrac%7Bn%28E_%7B2%7D%20%29%7D%7Bn%28S%29%7D%20%3D%20%5Cfrac%7B1%7D%7B6%7D)
But E₁ and E₂ are mutually exclusive events
P(E₁∩E₂) = 0
<u><em>Step(iii):-</em></u>
The probability that he will roll a 2 or a 3
we have P((E₁UE₂) = P(E₁)+P(E₂)
= ![= \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1%7D%7B6%7D%20%2B%20%5Cfrac%7B1%7D%7B6%7D%20%3D%20%5Cfrac%7B2%7D%7B6%7D%20%3D%20%5Cfrac%7B1%7D%7B3%7D)
![P((E_{1} UE_{2} ) = \frac{1}{3}](https://tex.z-dn.net/?f=P%28%28E_%7B1%7D%20UE_%7B2%7D%20%29%20%3D%20%5Cfrac%7B1%7D%7B3%7D)