Question

two dice are thrown simultaneously.find the probability that the sum of numbers appearing on their tops is less than or equal to 10​

Answer 1

Answer:

11/12...

Because:

Total number of outcomes = 36

Favorable outcomes with the sum as 11 are (5, 6) and (6, 5).

Favourable outcome with the sum as 12 is (6, 6).

P(sum 11) =  

P(sum 12) =  

P(sum 10) = 1 - [P(sum 11) + P(sum 12)] =1-1/12=11/12

Answer 2

Answer:

11/12=33/36

as we know that

sample space of two dice is

S={(1,1),(1,2)..........................................................................(6,6)]

n(S)=36

let a be the event that

number on the top is less than 10

the numbers that dont follow this law are

5,6

6,5

6,6

hence

n(A)=33

now for taking the probabilty

p(A)=n(A)/n(S)

=33/36

=11/12

Step-by-step explanation: