A and B throw a pair of dice. If A throws 9, find B’s chance of throwing a higher number.

When a pair of dice are thrown, then total no. of possible outcomes = 6 × 6 = 36, which are

{ (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) }

E ⟶ event of throwing a no. higher than 9.

No. of favourable outcomes = 6 {(4, 6) (5, 5) (6, 4) (5, 6) (6, 5) (6, 6)}

We know that P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes)

i.e., P(E) = 6/36 = 1/6

Given A pair of dice is thrown

Required to find: Probability that the total of numbers on the dice is greater than 9

First let’s write the all possible events that can occur

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6),

It’s seen that the total number of events is 6² = 36

Favourable events i.e. getting the total of numbers on the dice greater than 9 are

(5,5), (5,6), (6,4), (4,6), (6,5) and (6,6).

So, the total number of favourable events i.e. getting the total of numbers on the dice greater than 9 is 6

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting the total of numbers on the dice greater than 9 = 6/36 = 1/6