Question 1189432: I have 5 marbles numbered 1 through 5 in a bag. Suppose I take out two different marbles at random. What is the expected value of the product of the numbers on the marbles? Answer as a decimal to the nearest tenth.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
We have the numbers {1,2,3,4,5} in the bag.
We select 2 different marbles and multiply their face value.
Here's a 5x5 multiplication table mentioning the numbers above.
The color coding is to help distinguish the marble value (red or blue) and the stuff in black is the actual product.
| 1 | 2 | 3 | 4 | 5 | | 1 | 1 | 2 | 3 | 4 | 5 | | 2 | 2 | 4 | 6 | 8 | 10 | | 3 | 3 | 6 | 9 | 12 | 15 | | 4 | 4 | 8 | 12 | 16 | 20 | | 5 | 5 | 10 | 15 | 20 | 25 |
Since no replacement is happening, this means getting two marbles of the same number is impossible. So we'll ignore stuff along the main diagonal.
I'll replace those values with a horizontal line.
| 1 | 2 | 3 | 4 | 5 | | 1 | — | 2 | 3 | 4 | 5 | | 2 | 2 | — | 6 | 8 | 10 | | 3 | 3 | 6 | — | 12 | 15 | | 4 | 4 | 8 | 12 | — | 20 | | 5 | 5 | 10 | 15 | 20 | — |
The list of all possible outcomes is:
{2, 3, 4, 5, 6, 8, 10, 12, 15, 20}
This is the sample space. We only need to list each unique item once.
There are 20 items in the second table shown above. So the chances of getting a product of "8" for instance is 2/20 = 1/10 as there are two copies of "8" out of 20 items total.
Let X be the product and P(X) is the probability of having that product show up.
We can form this probability distribution table
| X | P(X) | | 2 | 1/10 | | 3 | 1/10 | | 4 | 1/10 | | 5 | 1/10 | | 6 | 1/10 | | 8 | 1/10 | | 10 | 1/10 | | 12 | 1/10 | | 15 | 1/10 | | 20 | 1/10 |
Each P(X) value is 1/10, so technically a table seems a bit overkill here and a bit silly. On the other hand, it's good practice to form such tables when it comes to expected value problems.
What we do from here is multiply each X and P(X) value together. Each row is handled separate.
We can extend that table to form a third column of X*P(X)
| X | P(X) | X*P(X) | | 2 | 1/10 | 2/10 | | 3 | 1/10 | 3/10 | | 4 | 1/10 | 4/10 | | 5 | 1/10 | 5/10 | | 6 | 1/10 | 6/10 | | 8 | 1/10 | 8/10 | | 10 | 1/10 | 10/10 | | 12 | 1/10 | 12/10 | | 15 | 1/10 | 15/10 | | 20 | 1/10 | 20/10 |
At this point, a table isn't such a bad idea to keep all the values organized.
I'm not reducing the fractions because adding them requires a common denominator
Add the X*P(X) values to get
(2/10)+(3/10)+(4/10)+(5/10)+(6/10)+(8/10)+(10/10)+(12/10)+(15/10)+(20/10)
(2+3+4+5+6+8+10+12+15+20)/10
85/10
8.5
The expected value is 8.5
If we were to do this experiment many many times (say 1000s of times), and average all the results, then we should get fairly close to 8.5
This empirical process of seeking out the expected value will not likely land exactly on 8.5 due to the random nature of this process. However, we should get fairly close to it.
Answer: 8.5
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