Answer:
The probability that he will roll a 2 or a 3
Ā [tex]P((E_{1} UE_{2} ) = \frac{1}{3}[/tex]
Step-by-step explanation:
Step(i);-
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
Step(ii):-
Let Eā be the event of roll a '2'
n(Eā) = 1
The probability that Dimas wants to roll a '2'
[tex]P(E_{1} ) = \frac{n(E_{1} )}{n(S)} = \frac{1}{6}[/tex]
Let Eā be the event of roll a '3'
n(Eā) = 1
The probability that Dimas wants to roll a '3'
[tex]P(E_{2} ) = \frac{n(E_{2} )}{n(S)} = \frac{1}{6}[/tex]
But Eā and Eā are mutually exclusive events
P(Eāā©Eā) = 0
Step(iii):-
The probability that he will roll a 2 or a 3
we have Ā P((EāUEā) = P(Eā)+P(Eā)
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā = [tex]= \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}[/tex]
Ā Ā Ā Ā Ā Ā [tex]P((E_{1} UE_{2} ) = \frac{1}{3}[/tex]
