Question

A box contains 30 tickets, bearing only one number from 1 to 30 on each. If one ticket is drawn at random, find the probability of an event that the ticket drawn bears
(1) an odd number
(2) a complete square number.

Answer 1

Total number of tickets = 30
(1)
Number of tickets bearing odd number = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29
= 15
Therefore, required probability = $\frac{\mathrm{Number}\mathrm{of}\mathrm{tickets}\mathrm{bearing}\mathrm{odd}\mathrm{number}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{tickets}}$
$$\begin{gathered} =\frac{15}{30} \\ =\frac{1}{2} \end{gathered}$$
Hence, the probability that the ticket drawn bears an odd number is $\frac{1}{2}$.
(2)
Number of tickets bearing perfect square number = 1, 4, 9, 16, 25
= 5
Therefore, required probability = $\frac{\mathrm{Number}\mathrm{of}\mathrm{tickets}\mathrm{bearing}\mathrm{perfect}\mathrm{number}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{tickets}}$
$$\begin{gathered} =\frac{5}{30} \\ =\frac{1}{6} \end{gathered}$$
Hence, the probability that the ticket drawn bears a perfect square number is $\frac{1}{6}$.