Find probability that the sum of two dice is not 6 and not 5?

Question

When rolling two fair dice, what is P(sum of two dice is not 6 and not 5)?

Calculation: First, I found the probability of two numbers that would roll a sum of 6:

(1,5) (2,4) (3,3) (4,2) (5,1) = 5/36 (each probability is 1/36)

Then, I found the probability of two numbers that would roll a sum of 5:

(1,4) (2,3) (3,2) (4,1) = 4/36

My first method was to add the probabilities and then subtract from one to give the probability of NOT rolling a sum of 6 or 5:

(5/36) + (4/36) = 9/36

1 - 9/36 = 27/36 = 0.75

This turned out to be the accepted answer of my online homework. But then, I realized that the original question asked for P(not sum of 6 AND not sum of 5).

So, I recalculated:

1 - 5/36 = 31/36

1 - 4/36 = 32/36

(31/36)(32/36) = 992/1296 ~ 0.7654 (using multiplication rule)

When I entered this fraction in, it was incorrect. But since the questions asks for "AND" and not "or", wouldn't the second probability be the actual correct answer?

Answer 1

By De Morgan's Laws:

not sum of 6 and not sum of 5 is the same to not (sum of 6 or sum of 5)

The second answer is not correct because "not sum of 6" and "not sum of 5" are not independent. So you cannot write the probability of their intersection as product of the individual probabilities (or as you called it "multiplication rule")

The first answer is correct because "sum of 6" and "sum of 5" are disjoint, and the probability of their union is the sum of the probabilities.

Answer 2

$P(A~\text{and}~B)$ i.e. $P(A\cap B)$ is equal to $P(A)\cdot P(B)$ ONLY WHEN THEY ARE INDEPENDENT EVENTS.

In general $P(A\cap B)\neq P(A)\cdot P(B)$

Instead, $P(A\cap B)=P(A)\cdot P(B\mid A)$

If you have no reason to expect the events to be independent, you must approach differently and your initial approach was indeed the correct one.

Answer 3

The probability that the sum of two dice is 6 and 5 is 9/36. So the probability that the sum of two dice is not 6 and 5 is 1- 9/36=27/36.