Statistics Probability Question?

Question

Please help me on my assignment, I'm having a hard time understanding it. Here are the work problems on probability:

Suppose you roll two fair dice.

Let A be the event the sum of the dice is 2, 3, or 4
Let B be the event the sum of the dice is 5, 6, 7, 8, 9, 10, 11, or 12
Let C be the event the sum of the dice is an even number
Let D be the event the first die rolled is a 2


a) Find P(A)
b) Find P(B) using the complement of the event, B.
c) Find P(C)
d) Find P(C or D)
f) Find P(C goven D)
g) Find P(D given C)

You don't have to answer all of it. I just wanted to see how you got some of the answers and if I'm also doing it correctly. Thanks!

Answer 1

There are 36 possibilities when rolling dice, not 11. (It's what's on the dice, not the sum that determines the possibilities.) They are:
Sum of 2 = 1-1
Sum of 3 = 1-2 or 2-1
Sum of 4 = 1-3, 3-1, or 2-2
Sum of 5 = 1-4, 4-1, 2-3, or 3-2
Sum of 6 = 1-5, 5-1, 2-4, 4-2, or 3-3
Sum of 7 = 1-6, 6-1, 2-5, 5-2, 3-4, or 4-3
Sum of 8 = 2-6, 6-2, 3-5, 5-3, or 4-4
Sum of 9 = 3-6, 6-3, 4-5, or 5-4
Sum of 10 = 4-6, 6-4, or 5-5
Sum of 11 = 5-6 or 6-5
Sum of 12 = 6-6

a. Count up how many of the possibilities have a sum of 2, 3, or 4. It's that many out of 35.

b. 1 - the answer to part A

c. Count up how many possibilities add up to 2, 4, 6, 8, 10, or 12. It's that many out of 36.

d. See how many of the odds have the first number "2" add that on to the number you used in part "c". ("Or" means +)

f. Find how many have the first number even (which should be 18). See how many of those have a sum that is even. Divide to get probability.

g. Look at all the sums that are even numbers. Count how many of those have the first number even. Divide to get probability.

Answer 2

Ok, when you roll two dice: there are 6 possible numbers for each dice. So the possible combinations are:

The sum of two dice:

1 2 3 4 5 6
+-------------------
1 | 2 3 4 5 6 7
2 | 3 4 5 6 7 8
3 | 4 5 6 7 8 9
4 | 5 6 7 8 9 10
5 | 6 7 8 9 10 11
6 | 7 8 9 10 11 12

So there are 36 different possible rolls. To find the probability of event A, you want to know how many possible rolls out of 36 will result in a sum of 2, 3, or 4.

If you look at the table, this happens 6 times. 6 times out of 36 or 6/36 which reduces to 1/6.

For Event B, the probability that the sum is 5-12, this is the complement of Event A because it is the probability that anything BUT A will happen. So, to find the complement of event A, you take 1 - P(A) = 1 - 1/6 = 5/6.

You could have also looked at the table and seen that a sum of 5-12 appears 30/36 times which reduces to 5/6. So the P(B) = 5/6 which is the same as the complement of Event A.

Event C would be a "success" if you roll an even sum, so if the sum is 2, 4, 6, 8, 10, 12. Looking again at the table, this happens 18 out of 36 times or 18/36.
So P(C) = 18/36 = 1/2.