SOLUTION: Suppose that we roll a pair of fair dice, so each of the 36 possible outcomes is equally likely. Let A denote the event that the first die lands on 3, let B be the event that the

Question 1190764: Suppose that we roll a pair of fair dice, so each of the 36 possible outcomes is equally likely. Let
A denote the event that the first die lands on 3, let B be the event that the sum of the dice is 8, and
let C be the event that the sum of the dice is 7.
a. Are A and B independent?
b. Are A and C independent?
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Part (a)

A = the first die lands on 3
B = the sum of the dice is 8

P(A) = 6/36 = 1/6 because there are 6 ways to roll the second die if the first die is fixed at "3", and this is out of 36 ways to roll two dice.

P(B) = 5/36 because there are five ways to sum to 8 as shown below
2+6 = 8
3+5 = 8
4+4 = 8
5+3 = 8
6+2 = 8

P(A and B) = 1/36
We have only one way to sum to 8 such that the first die is 3

Now notice that
P(A)*P(B) = (1/6)*(5/36) = 5/216
which does NOT match with P(A and B) = 1/36

So P(A and B) = P(A)*P(B) is a false statement

Therefore, the events A and B are not independent. We consider them dependent.

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Here's another way to see why A and B aren't independent.

We found earlier that
P(A) = 1/6

If somehow we know 100% that event B happened, then we have
P(A given B) = 1/5
this is because the sample space (originally 36 items) has reduced to 5 items only

If we know event B happened, then the sample space is
2+6 = 8
3+5 = 8
4+4 = 8
5+3 = 8
6+2 = 8
i.e. the five ways to add to 8. Of those 5 ways, only one has 3 as the first item.

P(A) = 1/6 doesn't match with P(A given B) = 1/5
We need P(A given B) = P(A) to be true if we want A and B to be independent.

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Part (b)

A = the first die lands on 3
C = the sum of the dice is 7

P(A) = 1/6 was calculated earlier.

Ways to sum to 7
1+6 = 7
2+5 = 7
3+4 = 7
4+3 = 7
5+2 = 7
6+1 = 7
There are 6 sums here out of 36 ways to roll two dice.
This means P(C) = 6/36 = 1/6

event A and C = first die lands on 3 AND the sum is 7
The only way for this to happen is to have the sum 3+4 = 7.
Which means,
P(A and C) = 1/36

Notice how
P(A)*P(C) = (1/6)*(1/6) = 1/36
which matches with the P(A and C)

We found that P(A and C) = P(A)*P(C) is a true equation.

Therefore, events A and C are independent.

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