A die is thrown. Find the probability of getting:

(i) a prime number

(ii) 2 or 4

(iii) a multiple of 2 or 3

Asked by Aaryan | 1 year ago |  33

##### Solution :-

Given: A die is thrown.

The total number of outcomes is six, n (S) = 6

(i) Let E be the event of getting a prime number

E = {2, 3, 5}

n (E) = 3

P (E) = $$\dfrac{ n (E) }{ n (S)}$$

$$\dfrac{3}{6}$$

$$\dfrac{1}{2}$$

(ii) Let E be the event of getting 2 or 4

E = {2, 4}

n (E) = 2

P (E) = $$\dfrac{ n (E) }{ n (S)}$$

$$\dfrac{2}{6}$$

$$\dfrac{1}{3}$$

(iii) Let E be the event of getting a multiple of 2 or 3

E = {2, 3, 4, 6}

n (E) = 4

P (E) = $$\dfrac{ n (E) }{ n (S)}$$

$$\dfrac{4}{6}$$

$$\dfrac{2}{3}$$

Answered by Aaryan | 1 year ago

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