Two standard dice are thrown. Determine the theoretical probability that the sum is a) 4 b) 7
c) an even number d) not a 6 e) not a perfect square

Step-by-step explanation:

there are 36 (6×6) possible results and number combinations for 2 dice.

a)

to create a sum of 4 we need one of the following combinations :

1 3

2 2

3 1

so, 3 out of the overall possible 36 possibilities.

and that means the probability is 3/36 = 1/12

b)

to create a sum of 7 we need one of the following combinations :

1 6

2 5

3 4

4 3

5 2

6 1

so, 6 out of the overall possible 36 possibilities.

and that means a probability of 6/36 = 1/6

c)

since the possible outcomes in that regard are only 2 (even and uneven), with equal balance the probability is 1/2.

make it more formal ?

to get an even sum, every possible number on one die can be combined with 3 possibilities on the other.

1 can be combined with 1, 3, 5

2 with 2, 4, 6

3 with 1, 3, 5

4 with 2, 4, 6

5 with 1, 3, 5

6 with 2, 4, 6

these are 6×3 = 18 combinations of of the possible 36.

so, the probability is 18/36 = 1/2

d)

not a 6 ?

so, what does that mean ?

we are looking only at 5 possible outcomes per die.

that is 5×5 = 25 combinations out of the possible 36.

that means that probabilty is 25/36.

e)

not a perfect square ?

we have in the range of the possible sums with 2 dice (2 .. 12) only two perfect squares : 4 and 9

so, the sum must NOT be 4 AND NOT be 9.

so, here it might be easier to count the unwanted cases and then deduct the probability from 1 to express the opposite.

the combinations to get 4 we have already under a)

3 combinations with 3/36=1/12 probability.

the combinations to get 9

1 none

2 none

3 6

4 5

5 4

6 3

so, 4 combinations with a probability of 4/36 = 1/9

therefore, we have (3+4) = 7 cases out of the possible 36 we actually want to avoid.

so, the probability to get any one of the other (desired) combinations is therefore (36-7)/36 = 29/36

it should be a.

hope it