Question

The probability of getting 11 when an ordinary die is thrown twice is......

Answer 1

Let E₇ = event of getting sum eaual to 11. The events of the sum of 11 will be E₇ = [(5, 6), (6, 5), ] = 2

Therefore, probability of getting ‘sum of 11’
p(E7)=number of favorable outcomes/total number of possible outcomes

Total number of possible outcomes=36 (Two different dice are thrown simultaneously being number 1, 2, 3, 4, 5 and 6 on their faces. We know that in a single thrown of two different dice, the total number of possible outcomes is (6 × 6) = 36)

= 2/36
= 1/18