Question

If a two digit number is chosen at random, then the probability that the number chosen is a multiple of 3, is
(a)$\frac{3}{10}$
(b)$\frac{29}{100}$
(c)$\frac{1}{3}$
(d)$\frac{7}{25}$

Answer 1

SOLUTION :

The correct option is (C)   : ⅓

Given : A two digit number is chosen at random  

Two digit numbers are from 10 to 99  

Number of two digit numbers = 99 - 10 + 1 = 90 (10 & 99 both inclusive)

Total number of possible outcome = 90

Let E = Event of getting a multiple of 3  

Multiple of 3  from 10 to 90  are = 12,15,18,........ 99

These numbers form an A.P  

So, a = 12 , d = 15 -12 = 3 , an = 99

an = a + (n - 1)d

99 = 12 + (n - 1) 3

99 = 12 + 3n - 3  

99 = 9 + 3n

99 - 9 = 3n

90 = 3n  

n = 90/3

n = 30  

Number of outcomes  favorable to E = 30

Probability ,P(E) = Number of favourable outcomes / total number of outcomes

P(E) = 30/90 = ⅓  

Hence, the Probability of getting a  Multiple of 3  ,P(E)  is 1/3.

HOPE THIS ANSWER WILL HELP  YOU….

Answer 2

Total elementary events=90 (10-90)

Multiples of 3 between them=30

So probability=Favourable elementary events/Total elementary events

=30/90

=1/3