Question

If you take a random 4 digit number, what is the probability that (i)the ten's place digit is greater than the unit's place digit
(ii) that the leftmost digit is a multiple of 3 (iii) the rightmost digit is a multiple of 3. [4 MARKS]

Answer 1

Concept : 1 Mark
Sub-parts : 1 Mark each

(i) There is a 1 in 10 chance that the ten’s place digit and one’s place digit are same. Therefore, there is a $\frac{9}{10}$ probability that these digits are different. In this latter case, it is equally likely for the ten’s place digit to be bigger or smaller than the unit’s place. Hence, the required probability is $\frac{9}{20}$.
(ii) The leftmost digit can take only the 9 values from 1 to 9. It cannot be zero. Out of these nine values, three are favourable. Hence the probability will be $\frac{3}{9}$ = $\frac{1}{3}$.
(iii) The rightmost digit can take any value from 0 to 9. Out of these 10 values, 3 are favourable. Hence, the probability is $\frac{3}{10}$ = 0.3.