Question

One lottery ticket is drawn at random from a bag containing 20 tickets numbered from 1 to 20. Find the probability that the number on the ticket drawn is
(i) either even or square of an integer
(ii) divisible by 3 or 5.

Answer 1

Let S be the sample space of the experiment.
S = {1, 2, 3, 4,..., 20}
∴ n(S) = 20

(i) Let A be the event that the number on the drawn ticket is even and B be the event that the number on the drawn ticket is the square of an integer.
A = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} and B = {1, 4, 9, 16}
Thus, we have:
A$\cap$ B = {4, 16}
Now,
n(A) = 10, n(B) = 4 and n(A$\cap$B) = 2
Thus, we get:
$$\begin{gathered} \mathrm{P}\left(\mathrm{A}\right)=\frac{n\left(\mathrm{A}\right)}{n\left(\mathrm{S}\right)}=\frac{10}{20}=\frac{1}{2} \\ \mathrm{P}(\mathrm{B})=\frac{n\left(\mathrm{B}\right)}{n\left(\mathrm{S}\right)}=\frac{4}{20}=\frac{1}{5} \end{gathered}$$
$P(\mathrm{A}\cap \mathrm{B})=\frac{n(\mathrm{A}\cap \mathrm{B})}{n\left(\mathrm{S}\right)}=\frac{2}{20}=\frac{1}{10}$
Now, on applying the addition theorem of probability, we get:
P(A $\cup$ B) = P(A) + P(B) $-$ P(A $\cap$ B)
$$\begin{gathered} =\frac{1}{2}+\frac{1}{5}-\frac{1}{10} \\ =\frac{5+2-1}{10}=\frac{6}{10}=\frac{3}{5} \end{gathered}$$
Therefore, the probability that the number on the drawn ticket is either even or square of a integer is $\frac{3}{5}$.

(ii) Let C be the event that the number on the drawn ticket is divisible by 3 and D be the event that the number on the drawn ticket is divisible by 5.
∴ C = {3, 6, 9, 12, 15, 18} and D = {5, 10, 15, 20}
Thus, we have:
C $\cap$ D = {15}
n(C) = 6, n(D) = 4 and n(C$\cap$ D) = 1
Now,

$$\begin{gathered} \mathrm{P}\left(\mathrm{C}\right)=\frac{n\left(\mathrm{C}\right)}{n\left(\mathrm{S}\right)}=\frac{6}{20}=\frac{3}{10} \\ \mathrm{P}\left(\mathrm{D}\right)=\frac{n\left(\mathrm{D}\right)}{n\left(\mathrm{S}\right)}=\frac{4}{20}=\frac{1}{5} \\ \mathrm{P}(\mathrm{C}\cap \mathrm{D})=\frac{n(\mathrm{C}\cap \mathrm{D})}{n\left(\mathrm{S}\right)}=\frac{1}{20} \end{gathered}$$
On applying the addition theorem of probability, we get:
P(C $\cup$ D) = P(C) + P(D) $-$ P(C $\cap$ D)
$$\begin{gathered} =\frac{3}{10}+\frac{1}{5}-\frac{1}{20} \\ =\frac{6+4-1}{20} \\ =\frac{9}{20} \end{gathered}$$
Therefore, the probability that the number on the drawn ticket is divisible by either 3 or 5 is $\frac{9}{20}$.