# Are these two dice rolls dependent or independent?

Summary:: Roll a dice twice.

Event A: First dice roll yield 2 or 5.
Event B: Sum of the two results are atleast 7

Is A and B independent?

If they are independent then
$$P(A \cap B) = P(A) * P(B)$$

P(A) = 2/6 = 1/3
P(B) = 1 - 9/36 = 27/36
$$P(A \cap B) = 7/36 \neq P(A)*P(B) = 1/4$$

Yet book says they are independent.

##P(B)## should just be a case of counting the number of outcomes that sum to greater or equal than 7, out of the possible 36 outcomes.

Homework Helper
Gold Member
2022 Award
P(B) = 1 - 9/36 = 27/36
Yet book says they are independent.
That P(B) is definitely not right.

One way to think about independence is whether the first event influences the second. In this case it appears that the probability of the second event depends on the first. But, does it? These two events look like they should be dependent, but are they really?

If you had to bet at the start on getting a total of ##7## or more, what would the probability be? And, if you were told that the first die was either a 2 or a 5, what would the revised probability be?

To put this formally, two events are independent if:
$$P(B|A) = P(B)$$
This says that ##B## is just as likely whether or not ##A## happens. Hence, alternatively:
$$P(A \cap B) = P(A)P(B|A) = P(A)P(B)$$

Last edited:
Hm, yea I calculated it to be 9 cases but there are 12 giving an outcome below 7.
P(B) =
1,5
1,4
1,3
1,2
1,1
2,1
2,2
2,3
2,4
3,1
3,2
3,3

P(B) = 1 - 12/36 = 24/36 = 2/3
P(A) * P(B) = 2/9

P(A & B) =
2,5
2,6
5,2
5,3
5,4
5,5
5,6
= 7/36

Would any of these have one less or one more combination then P(A&B) = P(A)P(B)
But as it is, there isnt.

Homework Helper
Gold Member
2022 Award
Hm, yea I calculated it to be 9 cases but there are 12 giving an outcome below 7.
P(B) =
1,5
1,4
1,3
1,2
1,1
2,1
2,2
2,3
2,4
3,1
3,2
3,3

You forgot:

4,1
4,2
5,1

• 