Question

If a digit is chosen at random from the digits 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

Total numbers of elementary events are = 9 

Let E be the event of getting a multiple of 3 

Favorable outcomes are: 3, 6, 9 

Numbers of favorable outcomes are: 3 

P (multiple of 3) = P (E) = \(\frac{3}9\) = \(\frac{1}3\)