Question

There are 15 tickets in a box, each bearing one of the numbers from 1 to 15. One ticket is drawn at random from the box. Find the probability of event that the ticket drawn -
(1) shows an even number.
(2) shows a number which is a multiple of 5.

Answer 1

Sample Space (S) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
∴ n(S) = 15

(1) Event A: the ticket shows an even number
A = {2, 4, 6, 8, 10, 12, 14}
∴ n(A) = 7
$$\begin{gathered} \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{n}\left(\mathrm{A}\right)}{\mathrm{n}\left(\mathrm{S}\right)} \\ =\frac{7}{15} \end{gathered}$$
Hence, the probability that the ticket shows an even number is $\frac{7}{15}$.
(2) Event B: the ticket shows a number which is a multiple of 5
B = {5, 10, 15 }
∴ n(B) = 3
$$\begin{gathered} \therefore \mathrm{P}\left(\mathrm{B}\right)=\frac{\mathrm{n}\left(\mathrm{B}\right)}{\mathrm{n}\left(\mathrm{S}\right)} \\ =\frac{5}{15} \\ =\frac{1}{5} \end{gathered}$$
​Hence, the probability that the ticket shows a number which is a multiple of 5 is $\frac{1}{5}$.