Question

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

Answer 1

SOLUTION :  

GIVEN: Two dice are thrown  

Here, two dice are thrown, so possible outcomes 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),

Hence, total number of outcomes = 36

Let E = Event of  getting the total of numbers on the dice is greater than 10

Here,the total of numbers on the dice greater than 10 are (5, 6), (6, 5) and (6, 6)

Number of outcomes favourable to E = 3

Probability (E) = Number of favourable outcomes / Total number of outcomes

P(E4) = 3/36 = 1/12

Hence, the probability of getting the total of numbers on the dice greater than 10 = 1/12 .

HOPE THIS ANSWER WILL HELP  YOU….

Answer 2

$\texbf{Answer}$

There are 6 face in a dice

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