math problem, i know it's late.:)?

Question

two dice are rolled and the product of the two numbers that appear is calculated, what is the expected value of a roll?

i will be back at school at 8am btw.i really did get alot of good help.

i will be back at school at 8am btw.i really did get alot of good help.

Answer 1

product: ways to produce it; frequency
1 : 1*1 ;......................... 1
2 : 1*2, 2*1 ;.................. 2 (3)
3 : 1*3, 3*1 ;.................. 2 (5)
4 : 1*4, 2*2, 4*1 ;........... 3 (8)
5 : 1*5, 5*1 ; ................. 2 (10)
6 : 1*6, 6*1, 2*3, 3*2 ; ... 4 (14)
7 : none ; ..................... 0
8 : 2*4, 4*2 ;.................. 2 (16)
9 : 3*3 ; ........................ 1 (17)
10: 2*5, 5*2 ; ................ 2 (19)
11: none
12: 2*6, 6*2, 3*4, 4*3 ;... 4 (23)
15: 3*5, 5*3 ;................. 2 (25)
16: 4*4 ; ....................... 1 (26)
18: 3*6, 6*3 ; ................ 2 (28)
20: 4*5, 5*4 ;................. 2 (30)
24: 4*6, 6*4 ;................. 2 (32)
25: 5*5 ; ....................... 1 (33)
30: 5*6, 6*5 ;................. 2 (35)
36: 6*6 ; ....................... 1 (36)

(the numbers in parenthesis is simply to check that we have 36 possibilities)

Now, expectation:
1*(1/36) + 2*(2/36) + 3*(2/36) + 4*(3/36) + 5*(2/36) + ... + 36*(1/36) = 441/36 = 12.25.
Now, that is hard work!


d:

Answer 2

Ok. You're right, it is late, but you're lucky that (us) mathematicians don't sleep all that much.
Now, here's what you need. You must construct a table/matrix, with 36 entries, i.e. it will have 6 columns and 6 rows, where the entry on the first row and first column will have value 1*1=1, the entry at the second row and first column will be 2*1=2 etc. In the end it should look like this:

1 2 3 4 5 6
2 4 6 8 10 12
3 6 9 12 15 18
4 8 12 16 20 24
5 10 15 20 25 30
6 12 18 24 30 36

Now, if you need to calculate, let's say for instance, the probability that the value you get is equal to 6 or less, then you go and count the number of entries in the matrix with value 6 or less. You should see that there are 5 of them. This means that your probability is that number divided by the total number of probabilities, aka 5/36.
I think that should do it. There are harder ways to do this, but this is the most easy and not so timely method.

Hope this helped you, good luck.

Answer 3

You can build up a table of 36 entries corresponding to the different outcomes:
1 . 2 . 3 . 4 . 5 . 6
2 . 4 . 6 . 8 .10 12
3 . 6 . 9 12 15 18
4 . 8 12 16 20 24
5 10 15 20 25 30
6 12 18 24 30 36

The sum of the first row is 21, and the other rows are just 2, 3, 4, 5, 6 times the first row, so the sum of all entries is 21 (1 + 2 + 3 + .. + 6) = 21^2 = 441. So the expected value is 441/36 = 49 / 4 = 12.25.

For two n-sided dice, it's pretty easy to show that the epected value is (n+1)^2 / 4.

Answer 4

Take your pick from among the following:
1 * 1 = 1
1 * 2 = 2
1 * 3 = 3
1 * 4 = 4
1 * 5 = 5
1 * 6 = 6
2 * 2 = 4
2 * 3 = 6
2 * 4 = 8
2 * 5 = 10
2 * 6 = 12
3 * 3 = 9
3 * 4 = 12
3 * 5 = 15
3 * 6 = 18
4 * 4 = 16
4 * 5 = 20
4 * 6 = 24
5 * 5 = 25
5 * 6 = 30
6 * 6 = 36

Answer 5

Late? It's only quarter to sizx on my clock.

The expected value is the sum of product of the value each outcome times its probability.

With two dice, there are 36 possible rolls, each with equal probability:
(1,1), (1,2), ... (1,6), (2,1), (2,2), ... (2,6), ..., (6,1), (6,2)...(6,6)

The corresponding products are:
1, 2, ...6, 2, 4, ...12, ..., 6, 12, ...36

So the expected value of the product is:

(1 + 2 + ... + 6 + 2 + 4 + ... + 12 + ... + 6 + 12 + ... 36) / 36

You can do the arithmetic.

Answer 6

Scarlet Manuka is right. However there is an easier way to do solve this, using the theorem which says that for independent random variables X and Y, E[X*Y] = E[X]*E[Y].

In this case let X and Y be the results for rolling the first die and the second die respectively. E[X] = E[Y] = (1 + ... + n)/n = (n+1)/2

So E[X*Y] = E[X]*E[Y] = [(n+1)/2]*[(n+1)/2] = [(n+1)^2]/4

Answer 7

expect a value from 1 - 36

the minimum value is 1 and the maximum is 36.

Answer 8

not sure what ur asking but i think you mean the most likely product:

i believe it to be 12

Answer 9

ok iam not a pro but arent u out of school but owell i think u should go to www.ask.com lol