Question

A box contains 36 tickets numbered from 1 to 36. One ticket is drawn at random. Find the probability that the number on the ticket is either divisible by 3 or is a perfect square.

Answer 1

Let S be the sample space of the random experiment.
∴ S = {1, 2, 3, 4,..., 36}
Or,
n(S) = 36
Now,
Let A be the event that the number on the ticket drawn is divisible by 3 and B be the event that the number on the ticket drawn is a perfect square.
∴ A = {3, 6, 9, 12, 15, 18, 21,24, 27, 30, 33, 36} and B = {1, 4, 9, 16, 25, 36}
$\Rightarrow$A $\cap$ B = {9, 36}
Now,
n(A) = 12, n(B) = 6 and n(A$\cap$ B) = 2

Thus, we have:
P(A) = $\frac{n\left(A\right)}{n\left(S\right)}=\frac{12}{36}=\frac{1}{3}$, P(B) = $\frac{n\left(B\right)}{n\left(S\right)}=\frac{6}{36}=\frac{1}{6}$ and P(A $\cap$ B) = $\frac{n(A\cap B)}{n\left(S\right)}=\frac{2}{36}=\frac{1}{18}$
Now, on applying the addition theorem of probability, we get:

P(A $\cup$ B) = P(A) + P(B) $-$ P(A $\cap$ B)

= $\frac{1}{3}+\frac{1}{6}-\frac{1}{18}=\frac{4}{9}$

$\therefore$ Probability that the number on the ticket is either divisible by 3 or is a perfect square = $\frac{4}{9}$