Question

One is asked to select a two -digit number. What is the probability of getting the first digit greater than the second digit ?

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{1}{5}$
• D. $\frac{1}{6}$

Answer 1

The correct option is B

$\frac{1}{2}$

The combination of numbers where the first digit is greater than the second digit are

10 ----------------------1

20 , 21 ---------2

30,31,32 ------------3

40 , 41, 42, 43---------------4

50 , 51, 52, 53, 54----------------5

60,61,62,63,64, 65----------- 6

70,71,72,73,74,75,76--------------7

80,81,82,83,84,85,86,87----------------8

90.91,92,93,94,95,96,97,98 -------------9

______________________________________________

Total Numbers = 45

Hence the total numbers where the first digit is greater than the second digit = 45

Total number of 2 digit number = 90

Hence the probability = $\frac{45}{90}$

=$\frac{1}{2}$