A box contains cords numbered from 1 to 20. A card is drawn at random from the box.
Question:

A box contains cords numbered from 1 to 20. A card is drawn at random from the box. Find the probability that the number on the drawn card is

(i) a prime number,

(ii) a composite number,

(iii) a number divisible by 3.

 

Solution:

There are 20 cards in the box. One card can drawn at random from the box in 20 ways.

∴ Total number of outcomes = 20

(i) The prime number cards in the box are 2, 3, 5, 7, 11, 13, 17 and 19. There are 8 prime number cards in the box. So, there are 8 ways to draw a card from the box which is a prime number card.

Favourable number of outcomes = 8

$\therefore P($ Number on the drawn card is a prime number $)=\frac{\text { Favourable number of outcomes }}{\text { Total number of outcomes }}=\frac{8}{20}=\frac{2}{5}$

(ii) The composite number cards in the box are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18 and 20. There are 11 composite number cards in the box. So, there are 11 ways to draw a card from the box which is a composite number card.

Favourable number of outcomes = 11

$\therefore \mathrm{P}($ Number on the drawn card is a composite number $)=\frac{\text { Favourable number of outcomes }}{\text { Total number of outcomes }}=\frac{11}{20}$

(iii) The cards with number divisible by 3 on them in the box are 3, 6, 9, 12, 15 and 18. There are 6 cards with number divisible by 3 on them in the box. So, there are 6 ways to draw a card from the box with number divisible by 3 on them.

Favourable number of outcomes = 6

$\therefore \mathrm{P}($ Number on the drawn card is a number divisible by 3$)=\frac{\text { Favourable number of outcomes }}{\text { Total number of outcomes }}=\frac{6}{20}=\frac{3}{10}$

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