Question

If digit is chosen at random from the digit 1, 2, 3, 4, 5, 6, 7, 8, 9, then , the probability that the digit is a multiple of 3 is
(a)$\frac{1}{3}$
(b)$\frac{2}{3}$
(c)$\frac{1}{9}$
(d)$\frac{2}{9}$

Answer 1

SOLUTION :  

The correct option is (a) :1/3 .

Given : Digits are chosen from 1, 2, 3, 4, 5, 6, 7, 8, 9  

Total number of outcomes = 9

Let E = Event of getting a multiple of 3

Multiples of 3 are = 3, 6, 9

Number of outcome favourable to E = 3

Probability (E) = Number of favourable outcomes / Total number of outcomes

P(E) = 3/9 = 1/3

Hence, the probability of getting a multiple of 3 , P(E) = 1/3

HOPE THIS ANSWER WILL HELP YOU ...

Answer 2

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so here we know that the multiples of 3 out of

1 2 3 4 5 6 7 8 9

are

3 6 9

so total events are

9

favourable events are

3
$here \: is \: your \: answer \\ \\ = \frac{3}{9} \\ \\ \\ \\ = \: \frac{1}{3} \\ \\ \\ hope \: it \: helps$

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