Two unbiased dice are thrown. Find the probability that the total of the numbers on the dice is greater than 10.

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

let E ⟶event of getting sum on dice greater than 10

then no. of favourable outcomes = 3 {(5, 6) (6, 5) (6, 6)}

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

i.e., P(E) = 3/36 = 1/12

Given: A pair of dice is thrown

Required to find: Probability that the total of numbers on the dice is greater than 10

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 10 are (5, 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 10 is 3.

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 10 = 3/36 = 1/12.

Possible outcomes, n(S) = 36 

Number of events of getting sum greater than 10, n(E) = 3

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{3}{36}\) = \(\frac{1}{12}\)