Question

If two balanced die are rolled, the possible outcomes can be represented as follows. (1, 1) (2, 1) (3, 1) (4, 1) (5, 1) (6, 1) (1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2) (1, 3) (2, 3) (3, 3) (4, 3) (5, 3) (6, 3) (1, 4) (2, 4) (3, 4) (4, 4) (5, 4) (6, 4) (1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5) (1, 6) (2, 6) (3, 6) (4, 6) (5, 6) (6, 6) determine the probability that the sum of the dice is 10.

Answer 1

3 of the 36 outcomes have a sum of 10. The probability is 3/36 = 1/12

Answer 2

Answer: $\dfrac{1}{12}$

Step-by-step explanation:

Given : If two balanced die are rolled, the possible outcomes can be represented as follows.

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

Counting all the outcomes , the total outcomes = 36

Outcomes having um of the dice is 10 = (5, 5) (4, 6)  (6, 4)

Number of outcomes having um of the dice is 10 = 3

i.e. Number of favorable outcomes = 3

We know that ,

$\text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total ouctomes}}$

Thus, the probability that the sum of the dice is 10 = $\dfrac{3}{36}=\dfrac{1}{12}$

Hence, the required probability = $\dfrac{1}{12}$