From 25 identical cards, numbered 1, 2, 3, 4, 5, ….. , 24, 25; one card is drawn at random. Find the probability that the number on the card drawn is a multiple of :

(i) 3

(ii) 5

(iii) 3 and 5

(iv) 3 or 5

We know that, there are 25 cards from which one card is drawn.

Number of possible outcomes = 25.

(i) From numbers 1 to 25, there are 8 numbers which are multiple of 3 i.e. {3, 6, 9, 12, 15, 18, 21, 24}

∴ Number of favourable outcomes = 8.

P(selecting a card with a multiple of 3) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{8}{25}$.

Hence, the probability of selecting a card with a multiple of 3 = $\dfrac{8}{25}$.

(ii) From numbers 1 to 25, there are 5 numbers which are multiple of 5 i.e. {5, 10, 15, 20, 25}

∴ Number of favourable outcomes = 5.

P(selecting a card with a multiple of 5) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{5}{25} = \dfrac{1}{5}$.

Hence, the probability of selecting a card with a multiple of 5 = $\dfrac{1}{5}$.

(iii) From numbers 1 to 25, there is only one number which is multiple of 3 and 5 i.e. {15}

∴ Number of favourable outcomes = 1.

P(selecting a card with a multiple of 3 and 5) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{25}$.

Hence, the probability of selecting a card with a multiple of 3 and 5 = $\dfrac{1}{25}$.

(iv) From numbers 1 to 25, there are 12 numbers which are multiple of 3 or 5 i.e. {3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25}

∴ Number of favourable outcomes = 12.

P(selecting a card with a multiple of 3 or 5) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{12}{25}$.

Hence, the probability of selecting a card with a multiple of 3 or 5 = $\dfrac{12}{25}$.