Question

If two dice are thrown what is the probability that the sum is
(1)greater than 8
(2)neither 7 nor 11

Answer 1

(1)

Solution:

All possible variants of dices are $6\cdot6=36$

Suitable variants of dices are $10$:

$(3,6)\\ (4,5),(4,6)\\ (5,4),(5,5),(5,6)\\ (6,3),(6,4),(6,5),(6,6)\\[0.2cm]$

Probability is $\dfrac{10}{36}=\dfrac{5}{18}$.

Answer: 5/18

(2)

Solution:

All possible variants of dices are $6\cdot6=36$

Variants of dices which sum is 7 are $6$:

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

Variants of dices which sum is 11 are $2$:

$(5,6),(6,5)$

Suitable variants of dices are all except 7 and 11:

$36-6-2=28\\[0.2cm]$

Probability is $\dfrac{28}{36}=\dfrac{7}{9}$.

Answer: 7/9