6. A pair of fair dice are rolled and the sum of the numbers on the top faces is noted.
a) There are two probabilistic models that we have used for the experiment of rolling a pair of dice. The first one is the 36 outcome equally likely probabilistic model which is based on the fact that if the dice are fair, then any face is as likely to come up as any other face on each die. The multiplication principle says that since there are 6 faces on each die, there are 36 total possible outcomes. They can be listed in a table as
This model can be streamlined a bit in this problems because we are only interested in the sum of the number on the top faces, not the individual numbers making up the sum. We see from the body of the table that the outcomes can be sumarized as the numbers between 2 and 12, and the probability of each number is the number of times that number appears in the table divided by 36.
n |
p |
2 |
1/36 |
3 |
2/36 |
4 |
3/36 |
5 |
4/36 |
6 |
5/36 |
7 |
6/36 |
8 |
5/36 |
9 |
4/36 |
10 |
3/36 |
11 |
2/36 |
12 |
1/36 |
totals |
36/36 |
b) To compute the probability that the number is even, add up all the probabilities of all the outcomes in the event
E = {2, 4, 6, 8, 10, 12}
On the one hand, this is not too surprising, but on the other hand, it is. It sounds pretty reasonable that the probability if rolling an even number would be 1/2, but if you look at the model, there are six even numbers, but only 5 odd numbers. (It is impossible to roll a 1 with a pair of dice. Moreover, the even numbers all have odd numerators in their probabilities,and the odd numbers all have even numbers in their probabilities. There are quite a few influences which have to cancel out in order for the probability of rolling an even number to come out to 1/2, but they do.
You will also find that the probability that the number is divisible by 3 is 1/3, the probability that it is divisible by 4 is 1/4, and the probability that it is divisible by 6 is 1/6. One might be lead to conjecture that the probability that a number is divisible by n is 1/n.The next problem provides us with the first counterexample to that conjecture.
c) To compute the probability that the number is divisible by 5, add up all the probabilities of all the outcomes in the event
F = {5, 10}
7/36 is not 1/5. We could tell that the probability will not be 1/5, because a common denominator of the probabilities of alll the outcomes in the event is 36, and there is no fraction whose denominator is 36 which owill reduce to 1/5. However, it comes as close as a fraction whose denominator goes into 36 can come. You willalso find that the conjecture is also false for any number larger than 6.
d) To get the conditional probability that it is divisible by 5 given that it is even, we use the formula for the definition of conditional probability.
In this case, the intersection of E and F is {10}, so we get
e) 1/6 is just slightly less than 7/36, so the events are not independent. If we know that the number is even, that makes it a little less likely that the number will be divisible by 5.