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. What is the probability that the number on the card is divisible by 2, 3, or both?

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{2}{5}$
• D. $\frac{2}{3}$

Answer 1

The correct option is D $\frac{2}{3}$

Given,
24 are cards numbered 1, 2, 3, ..., 23, 24.
Here, the total number of possible outcomes = 24
Numbers divisible only by 2 are 2, 4, 8, 10, 14, 16, 20, and 22.

Therefore, 8 numbers are divisible only by 2 ...(i)
Numbers divisible only by 3 are 3, 9, 15, and 21.
Therefore, 4 numbers are divisible only by 3 ...(ii)
Numbers divisible by 2 and 3 both are 6, 12, 18, and 24.
Therefore, 4 numbers are divisible by 2 and 3 both ...(iii)

From (i), (ii), and (iii), we get:
The number of favorable outcomes = 8 + 4 + 4 = 16
We know that the probability of an event, E is given by,
$P\left(E\right)=\frac{\text{number of favorable}}{\text{total number of outcomes}}$

$\Rightarrow P\left(E\right)=\frac{16}{24}=\frac{2}{3}$

Therefore, the probability that the number on the card is divisible by 2, 3, or both is $\frac{2}{3}$.