Question

24 cards numbered 1, 2, 3, ...., 23, 24 are put in a box and mixed thoroughly. One person draws a card from the box. Find the probability that the number on the card is divisible by 2 or 3 or both.

Answer 1

Given,
24 cards numbered 1, 2, 3, ...., 23, 24
Here, the total possible outcomes = 24

Numbers divisible by only 2 are 2, 4, 8, 10, 14, 16, 20, 22
$\therefore$ 8 numbers are divisible by only 2 ...(i)

Numbers divisible by only 3 are 3, 9, 15, 21
$\therefore$ 4 numbers are divisible by only 3 ...(ii)
Numbers divisible by both 2 and 3 are 6, 12, 18, 24
$\therefore$ 4 numbers are divisible by both 2 and 3 ...(iii)

From (i),(ii) and (iii), we get
The number of favourable outcomes = 8 + 4 + 4 = 16
We know that, Probability of an event $E$ is given by,
$\text{P(}E\text{) =}$$\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
$\Rightarrow$$\text{P(}E\text{)=}$ $\frac{16}{24}$
$=\frac{2}{3}$
$\therefore$ The probability that the number on the card is divisible by 2 or 3 or both is $\frac{2}{3}$