word problem

Question

Two fair dice are tossed twice. Find the probability that the sum is 10 or less on the first toss and the sum is 8 or more on the second toss. What is the probability of five or more on the first toss or 8 or more on the second toss? Are these independent (why)? are they mutually exclusive (why)?

Answer 1

Ray,

 

On your last equation your calculation (for the second question) of the "overlap" is not correct. Technically there is no overlap since they are separate events.

 

You are looking for the probability that both events "occur" and since they are independent you multiply (30/36)(15/36).

 

The formula you are trying to use is P(A or B)=P(A) + P(B) - P(A and B)

 

which becomes: P(A or B) = P(A) + P(B) - P(A)P(B)

 

                        so we get  5/6  +  5/12  -  25/72           

Answer 2

First, we should acknowledge that, when two dice are rolled, there are 36 different possible outcomes (6*6 = 36).

 

Because there are many outcomes that result in a sum of 10 or less, let's consider the possible outcomes in which the sum is 11 or 12. The outcomes would be: (5,6), (6,5), and (6,6). There are only 3. This means that there are 33 outcomes in which the sum is 10 or less. So, the probability that the sum is 10 or less is (33/36).

For the sake of the rest of the problem, I will outline the number of outcomes for each sum.

 

Sum of 2: 1

Sum of 3: 2

Sum of 4: 3

Sum of 5: 4

Sum of 6: 5

Sum of 7: 6

Sum of 8: 5

Sum of 9: 4

Sum of 10: 3

Sum of 11: 2

Sum of 12: 1

 

So, for a sum of 8 or more, there are (5 + 4 + 3 + 2 +1) 15 possibilities. Then the probability of rolling a sum of 8 or more is (15/36).

Because the first question is an "and" probability, we multiply the probabilities of each roll.

 

                    (33/36)*(15/36) = .382 (rounded decimal) 

 

The number of outcomes that result in a sum of 5 or more is (4 + 5 + 6 + 5 + 4 + 3 + 2 + 1) 30. So, the probability that that the sum is 5 or more is (30/36).

 

We know from the first question that the probability of rolling a sum of 8 or more is (15/36).

 

Now, what we have here is bit of overlap. Included in both a sum of 5 or more and a sum of 8 or more are the sums of 8, 9, 10, 11, and 12. So, we also need the probability of the overlap cases. The amount of outcomes that have a sum of 8, 9, 10, 11, and 12 is (5 + 4 + 3 + 2 + 1) 15. Thus a probability of (15/36).

 

Because the second question is an "or" probability, we add the probabilities of each roll. But, because the also have overlap, we must also subtract the probability of the overlap.

             (30/36) + (15/36) - (15/36) = (30/36) = (5/6).

 

I believe both of these rolls are independent as what is rolled on the second roll is not dependent on what is rolled on the first. However, the second "or" question is not mutually exclusive as there are overlapping cases. 

I apologize if this is jumbled. Statistics and Probability is not my strongest. 

 

This might help: http://www.regentsprep.org/regents/math/algebra/apr8/LProbAO.htm

 

-Ray Biggerstaff