Question

Dimas is playing a game that uses a number cube. The sides of the cube are labeled from 1 to 6. Dimas wants to roll a 2 or a 3. He rolls the
number cube one time,
What is the probability that he will roll a 2 or a 3?

Who ever answer FIRST I WILL GIVE YOU BRAINIEST ANSWER

Answer 1

Answer:

The probability that he will roll a 2 or a 3

 $P((E_{1} UE_{2} ) = \frac{1}{3}$

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'

$P(E_{1} ) = \frac{n(E_{1} )}{n(S)} = \frac{1}{6}$

Let E₂ be the event of roll a '3'

n(E₂) = 1

The probability that Dimas wants to roll a '3'

$P(E_{2} ) = \frac{n(E_{2} )}{n(S)} = \frac{1}{6}$

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)

                               = $= \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$

            $P((E_{1} UE_{2} ) = \frac{1}{3}$