BITSAT 2022 Paper

Let P be the probability of getting an even number. Then by hypothesis the probability of getting an odd number is 3P. Since the events of getting an even number and an odd number are mutually exclusive and exhaustive.
∴P+3P=1⇒P=

1

4

Thus, the probability of getting an odd number in a single throw is

3

4

and that of an even number is

1

4

.
If the die is thrown twice, then the sum of the numbers in two throws is even if both the numbers are even or both are odd.
∴ Required probability =

3

4

×

3

4

+

1

4

×

1

4

=

10

16

=

5

8

.