# probability distribution and expected value

• May 4th 2010, 02:05 PM
bhuang
probability distribution and expected value
I have attempted the question, but I am doubting my answer because it is a rational number and the numbers in context are not.

I am providing the whole question for context:
Two standard six-sided dice are tossed. Let X be the sum of the scores on the two dice.
(a) Find
(i) P(X=6)
(ii) P (X>6)
(iii) P(X=7 | X>5)

(b) Elena plays a game where she tosses two dice.
If the sum is 6, she wins 3 points.
If the sum is greater than 6, she wins 1 point.
If the sum is less than 6, she loses kpoints.

Find the value of k for which Elena's expected number of points is zero.

I need verification with part b).
What I did was construct a probability distribution table where X was k, 3, and 1. And my probabilities were 10/36, 5/36, 21/36, respectively. Then I set the expected value equal to zero:
(10k/36) + (3 x 5/36) + (1 x 21/36) =0
to solve for k.

I get k as -36/10.

But I'm not sure if this is right, can someone please check if my method is correct?
• May 4th 2010, 02:11 PM
Bushy
Quote:

Originally Posted by bhuang

I am providing the whole question for context:
Two standard six-sided dice are tossed. Let X be the sum of the scores on the two dice.
(a) Find
(i) P(X=6)
(ii) P (X>6)
(iii) P(X=7 | X>5)

You should make a two row table. first row $x= 2,3\dots ,12$ with a second row being $P(x)$ there will be some symmetry here.
• May 4th 2010, 03:03 PM
bhuang
My probability distribution table is my two row table. Instead of 2, 3, ... I put the points obtained from the game in. It should be the same thing, right? What do you mean by symmetrical...
• May 4th 2010, 03:13 PM
Bushy
Quote:

Originally Posted by bhuang
My probability distribution table is my two row table. Instead of 2, 3, ... I put the points obtained from the game in. It should be the same thing, right? What do you mean by symmetrical...

The sum of two dice can only be from 2 to 12. These are the x's.

You then need to find the probabililty of each of these occuring.

I.e $P(2) =\frac{1}{36}, P(3) =\frac{2}{36}, P(4) =\frac{3}{36}, \dots ,P(12) =\frac{1}{36}$

Do you know how I found these? After you have this information you can answer the questions.