Question

A box contains cards bearing numbers from 6 to 70. If one card is drawn at random from the box, find the probability that it bears

(i) a one digit number.

(ii) a number divisible by 5.

(iii) an odd number less than 30.

(iv) a composite number between 50 and 70.

Answer 1

(i) Let E be the event of getting a one digit number.

Number of possible outcomes = 70 − 6 + 1 = 65

The outcomes favourable to E are 6, 7, 8 and 9.

\[\therefore\] Number of favourable outcomes = 4

\[P \left( E \right) = P \left( \text{Getting a one digit number} \right) = \frac{4}{65}\]

(ii) Let F be the event of getting a number divisible by 5.

Number of possible outcomes = 65

The outcomes favourable to the event F are 10, 15, 20, ..., 65, 70.

∴∴ Number of outcomes favourable to F = 13

\[P \left( F \right) = P \left( \text{Getting a number divisible by 5} \right) = \frac{13}{65} = \frac{1}{5}\]

(iii) Let G be the event of getting an odd number less than 30.

Number of possible outcomes = 65

The outcomes favourable to the event G are 7, 9, 11, 13, ..., 29.

\[\therefore\] Number of favourable outcome = 12

\[P \left( G \right) = P \left( \text{Getting an odd number less than 30} \right) = \frac{12}{65}\]

(iv) Let H be the event of getting a composite number between 50 and 70.

Number of possible outcomes = 65

The outcomes favourable to the event H are 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68 and 69.

\[\therefore\] Number of favourable outcomes = 15

\[P \left( H \right) = P \left( \text{Getting a composite number between 50 and 70} \right) = \frac{15}{65} = \frac{3}{13}\]