The probability that a two digit number selected at random will be a multiple of ‘3’ and not a multiple of ‘5’ is

Question

The probability that a two digit number selected at random will be a multiple of ‘3’ and not a multiple of ‘5’ is (A) (2/15) (B) (4/15) (C) (1/15) (D) (4/90)

Tardigrade Admin · Accepted Answer

24 out of the 90 two digit numbers are divisible by â3â and not by â5â. The required probability is therefore, (24/90) = (4/15)

Q. The probability that a two digit number selected at random will be a multiple of ‘3’ and not a multiple of ‘5’ is

$\frac{2}{15}$

$\frac{4}{15}$

$\frac{1}{15}$

$\frac{4}{90}$

24 out of the 90 two digit numbers are divisible by ‘3’ and not by ‘5’.
The required probability is therefore,
$\frac{24}{90} = \frac{4}{15}$

Q. The probability that a two digit number selected at random will be a multiple of ‘3’ and not a multiple of ‘5’ is

152​

154​

151​

904​

24 out of the 90 two digit numbers are divisible by ‘3’ and not by ‘5’. The required probability is therefore, 9024​=154​

$\frac{2}{15}$

$\frac{4}{15}$

$\frac{1}{15}$

$\frac{4}{90}$

24 out of the 90 two digit numbers are divisible by ‘3’ and not by ‘5’.
The required probability is therefore,
$\frac{24}{90} = \frac{4}{15}$