Question

a cubical die number from 1 to 6 is rolled twice find the probability of getting the sum of number on its faces is 10​

Answer 1

When a cubical die numbered from 1 to 6 is rolled twice, there are a total of 36 combination possible.

These possible combination are as follows:

{

(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)

}

Total number of combination, $n(S) = 36$

Let A be the event that sum of number on its faces is 10.

The number of possible combination, $n(A) = 3$ i.e. {(4,6),(6,4),(5,5)}

Formula for Probability is:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

Probability for the event A:

$P(A) = \dfrac{n(A)}{n(S)}$

$P(A) = \dfrac{3}{36}\\P(A) = \dfrac{1}{12}$

So, the required probability is $\frac{1}{12}$.