# Distribute, Expect!

• Aug 4th 2010, 08:03 AM
Aeonitis
Distribute, Expect!
Got a simple question i am trying to solve guys, could definitely do with some help. I am very glad my question will come easy to you guys.

Two numbers are chosen at random from {1,2,3} with repetitions allowed. Let the random variable X denote the sum of the 1st number and 2 times the 2nd number.

(i) Find the Distribution of X.
(ii) Find the Expectation E(X) of X.

Thank you, bad math student trying to catch up (Headbang)
• Aug 4th 2010, 09:55 AM
SpringFan25
I cant see a clever method to do this, so i'd do it the old fashioned way.

there are only 18 possible outcomes, each with probability 1/18.

Code:

Chosen    |  X  | Prob
1,1,1    |  2  | 1/18
1,1,2    |  4  | 1/18
1,1,3    |  6  | 1/18
2,1,1    |  2  | 1/18
..... etc

As an example, from the incomplete table above, we have:

P(X=2) = 2/18
(it appears twice, each with probability 1/18)

When you do the full table it will appear more times.

Use the same method to work out the chance of X taking all possible values.
• Aug 5th 2010, 02:34 AM
Aeonitis
Quote:

Originally Posted by SpringFan25
...we have:

P(X=2) = 2/18
(it appears twice, each with probability 1/18)

"Let the random variable X denote the sum of the 1st number and 2 times the 2nd number." I would guess that the first value of X would be '3' since the first number is one, added with 2 times 1 with a result of two giving 3 as an answer.

I sound like a smartass, but i am getting nowhere >_<
• Aug 5th 2010, 02:50 AM
Quote:

Originally Posted by Aeonitis
Got a simple question i am trying to solve guys, could definitely do with some help. I am very glad my question will come easy to you guys.

Two numbers are chosen at random from {1,2,3} with repetitions allowed. Let the random variable X denote the sum of the 1st number and 2 times the 2nd number.

(i) Find the Distribution of X.
(ii) Find the Expectation E(X) of X.

Thank you, bad math student trying to catch up (Headbang)

(1,1)...1+2(1)=3
(2,1)...2+2(1)=4
(3,1)...3+2(1)=5
(1,2)...1+2(2)=5
(2,2)...2+2(2)=6
(3,2)...3+2(2)=7
(1,3)...1+2(3)=7
(2,3)...2+2(3)=8
(3,3)...3+2(3)=9

Finally E(X) is the mean or average of X (the sums)
• Aug 5th 2010, 04:17 AM
Aeonitis
Awesome Archie, and that is the full answer, THAT SIMPLY?! oh, just making sure, i mean the right answer would be that the expectation of X i.e. E(X) is 54/9 giving a result of 6?!
• Aug 5th 2010, 04:59 AM