Tickets numbered 2, 3, 4, 5, ..., 100, 101 are placed in a box and mixed thoroughly.

Question:

Tickets numbered 2, 3, 4, 5, ..., 100, 101 are placed in a box and mixed thoroughly. One ticket is drawn at random from the box. Find the probability that the number on the ticket is
(i) an even number
(ii) a number less than 16
(iii) a number which is a perfect square
(iv) a prime number less than 40

Solution:

All possible outcomes are  2, 3, 4, 5................101.
Number of all possible outcomes = 100

(i) Out of these the numbers that are even = 2, 4, 6, 8...................100
Let E1 be the event of getting an even number.

Then, number of favourable outcomes = 50​   $\left[T_{n}=100 \Rightarrow 2+(n-1) \times 2=100, \Rightarrow n=50\right]$

$\therefore P$ (getting an even number) $=\frac{50}{100}=\frac{1}{2}$

(ii) Out of these, the numbers that are less than 16 = 2,3,4,5,..................15.
Let E2 be the event of getting a number less than 16.
Then,​ number of favourable outcomes =​ 14

$\therefore P$ (getting a number less than 16 ) $=\frac{14}{100}=\frac{7}{50}$

(iii) Out of these, the numbers that are perfect squares =  4, 9,16,25, 36, 49, 64, 81 and 100
Let E3 be the event of getting a number that is a perfect square.

Then,​ number of favourable outcomes = 9
$\therefore P$ (getting a number that is a perfect square) $=\frac{9}{100}$

(iv) Out of these, prime numbers less than 40 = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37
Let E4 be the event of getting a prime number less than 40.
Then,​ number of favourable outcomes =​ 12
$\therefore P$ (getting a prime number less than 40$)=\frac{12}{100}=\frac{3}{25}$