Question:
Someone is asked to take a number from 1 to 100. The probability that it is a prime,is
(a) $\frac{1}{5}$
(b) $\frac{6}{25}$
(c) $\frac{1}{4}$
(d) $\frac{13}{50}$
Solution:
(c) Total numbers of outcomes $=100$
So, the prime numbers between 1 to 100 are $2,3,5,7,11,13,17,19,23,29,31,37,41.43,47,53,56,61,67,71$,
$73,79,83,89$ and 97 .
$\therefore$ Total number of possible outcomes $=25$
$\therefore$ Required probability $=\frac{25}{100}=\frac{1}{4}$