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Sagot :
Answer:
The probability that he will roll a 2 or a 3
[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 E₁ be the event of roll a '2'
n(E₁) = 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 E₂ be the event of roll a '3'
n(E₂) = 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 E₁ and E₂ are mutually exclusive events
P(E₁∩E₂) = 0
Step(iii):-
The probability that he will roll a 2 or a 3
we have P((E₁UE₂) = P(E₁)+P(E₂)
= [tex]= \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}[/tex]
[tex]P((E_{1} UE_{2} ) = \frac{1}{3}[/tex]
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