Answer:

there are no probably to nor get 7 or 11 in pair of dice

Answer:

7/9

Step-by-step explanation:

Total number of all possible outcomes, n(S)=6×6=36

Let A: be the event that the sum is 7.

Hence the favourable outcomes are {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}, i.e., n(A)=6

Hence the probability of getting a total of 7. P(E)=

n(S)

n(A)

=

36

6

Similarly, let B: be the event that the sum is 11

Hence the favourable outcomes are {(5, 6), (6, 5)}, i.e., n(E)=2

Hence the probability of getting a total of 11 P(B)=

n(S)

n(B)

=

36

2

Since the events getting sum7 and getting sum 11 are independent of each other and mutually exclusive. Hence, P(A∩B)=0

Now, the probability of getting a total of 7 or 11 or both is

P(A∪B)=P(A)+P(B)−P(A∩B)

⟹P(A∪B)=

36

6

+

36

2

=

36

8

Therefore, the probability that the sum is neither 7 nor 11 is

P(A∪B)

′

=1−P(A∪B)=1−

36

8

=

36

28

=

9

7