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# Expectation theory (S1) watch

1. Hi,

I've been doing questions on expectation theory, and am stuck on one.

"Y is the larger score showing when two dice are thrown. Calculate E(Y)."

Can anyone help?

Thanks
2. Firstly, you need to find the probability that Y = 1,2,3,4,5,6. To do this, I think it is easiest to draw a table showing the different values possible on each dice, then fill in the larger score for each option, like this:

- 1 2 3 4 5 6

1 1 2 3 4 5 6
2 2 2 3 etc
3
4
5
6

Then the probability that y=5 is the number of fives in the table over 36.

Draw out a table to show P(Y=y) for values of y 1,2,3,4,5,6 and then
3. (Original post by sonofdot)
Firstly, you need to find the probability that Y = 1,2,3,4,5,6. To do this, I think it is easiest to draw a table showing the different values possible on each dice, then fill in the larger score for each option, like this:

1 2 3 4 5 6

1 1 2 3 4 5 6
2 2 2 3 etc
3
4
5
6

Then the probability that y=5 is the number of fives in the table over 36.

Draw out a table to show P(Y=y) for values of y 1,2,3,4,5,6 and then
Thanks, if I do that, and add up all the probabilities, I get 20/36 though, as there are a lot of numbers that are above 6 in the table..

is that okay?
4. (Original post by Darkest Knight)
Thanks, if I do that, and add up all the probabilities, I get 20/36 though, as there are a lot of numbers that are above 6 in the table..

is that okay?
Really? The probabilities should (and do) add up to 1

If you continue filling up the table, a bit of a pattern emerges:

- 1 2 3 4 5 6
1 1 2 3 4
2 2 2 3 4
3 3 3 3 4
4 4 4 4 4 etc
5. I don't understand how your working out the table. I thought you multiply by the number and the top by number at left...
6. (Original post by Darkest Knight)
I don't understand how your working out the table. I thought you multiply by the number and the top by number at left...
The numbers on the top are the scores of the first dice. The numbers down the left are the scores of the second dice. If the score of the first dice is 2 and the score of the second dice is 1, then the larger score (Y) is 2, so 2 goes in that box. If the score on both dice is 4, then the larger score is 4, so 4 goes in that box

Sorry if that just confused you more!
7. (Original post by sonofdot)
The numbers on the top are the scores of the first dice. The numbers down the left are the scores of the second dice. If the score of the first dice is 2 and the score of the second dice is 1, then the larger score (Y) is 2, so 2 goes in that box. If the score on both dice is 4, then the larger score is 4, so 4 goes in that box

Sorry if that just confused you more!
Okay, thanks that makes sense!

Also, please could you tell me, if I have a probability distribution, and don't have with x = 1 - 5, but then only having 3 of the values of P(X=x), so missing two (a and b) how can I work out those two values? I have the value of E(X)..

Thanks
8. (Original post by Darkest Knight)
Okay, thanks that makes sense!

Also, please could you tell me, if I have a probability distribution, and don't have with x = 1 - 5, but then only having 3 of the values of P(X=x), so missing two (a and b) how can I work out those two values? I have the value of E(X)..

Thanks
This becomes a simultaneous equation:

the sum of the probabilites (a + b + the ones you are given) = 1
And use the formula I gave above for E(X) using the value of E(x) given to get another equation

Incidentally, where is this equation from? I seem to recall doing it at some point last year... it isn't perhaps an OCR S1 exam question, is it?
9. (Original post by sonofdot)
This becomes a simultaneous equation:

the sum of the probabilites (a + b + the ones you are given) = 1
And use the formula I gave above for E(X) using the value of E(x) given to get another equation

Incidentally, where is this equation from? I seem to recall doing it at some point last year... it isn't perhaps an OCR S1 exam question, is it?
from my textbook thanks for your help
10. Sorry one last thing.. (and +rep given by the way)

How can I find the probability that X > E(X)?

Thanks!
11. (Original post by Darkest Knight)
Sorry one last thing.. (and +rep given by the way)

How can I find the probability that X > E(X)?

Thanks!
Have you worked out the value of E(X)?

If so, the probability that X > E(X) = sum of all probablilites for values > E(X)

For example, if E(X) = 4.6, then P(X > E(X)) = P(X=5) + P(X=6) + P(x=7)...
12. (Original post by sonofdot)
Have you worked out the value of E(X)?

If so, the probability that X > E(X) = sum of all probablilites for values > E(X)

For example, if E(X) = 4.6, then P(X > E(X)) = P(X=5) + P(X=6) + P(x=7)...
Sorry, really confused there.

I have the value of E(X) which is 7.4

X= 4, 6, 7, 10, 11
P(X=x) = 0.2, 0.175, 0.3, 0.125, 0.2

Thanks
13. (Original post by Darkest Knight)
Sorry, really confused there.

I have the value of E(X) which is 7.4

X= 4, 6, 7, 10, 11
P(X=x) = 0.2, 0.175, 0.3, 0.125, 0.2

Thanks
Then P(X > E(X)) = P(X=10) + P(X=11)
14. (Original post by sonofdot)
Then P(X > E(X)) = P(X=10) + P(X=11)
The books says the answer is 0.325!
15. (Original post by Darkest Knight)
The books says the answer is 0.325!
P(X=10) = 0.125
P(X=11) = 0.2

P(X>E(X)) = P(X=10) + P(X=11) = 0.125 + 0.2 = 0.325
16. (Original post by sonofdot)
P(X=10) = 0.125
P(X=11) = 0.2

P(X>E(X)) = P(X=10) + P(X=11) = 0.125 + 0.2 = 0.325
Ah, I'm still confused.

How does P(X=10) = 0.125

Bit confused where the "X =" comes from too..

Sorry!
17. (Original post by Darkest Knight)
Ah, I'm still confused.

How does P(X=10) = 0.125

Bit confused where the "X =" comes from too..

Sorry!
Ok - I'm not the best at explaining things, but I'll do my best...

X is a discrete random variable, something that can take different values, like the number on a dice

P(X=a) means the probability that X takes a certain value a; for example on a dice P(X=5) is the probability that you roll a 5, which is 1/6

E(X) is an average value for X; if you rolled the dice loads of times and worked out the mean score, E(X) is what you'd expect it to be

So E(X) is just like another value for X

In the problem you posted above, the only values that X can take are 4, 6, 7, 10 and 11

E(X) is 7.4: X obviously can't be exactly 7.4, but it can be some values bigger than 7.4. The only values it can take bigger than 7.4 are 10 and 11

So the probability that X>E(X) = the probability that takes a value bigger than E(X), and the only values of X for which this is possible are 10 and 11

so P(X>E(X)) = P(X = 10 or 11) = P(X = 10) + P(X=11)

From the probability table you posted, P(X=10) = 0.125 and P(X=11) = 0.2, which is where that came from

Sorry for the long-winded post, but maybe you understand a bit better now?

Sorry for the time delay too - I was eating tea!
18. (Original post by sonofdot)
Ok - I'm not the best at explaining things, but I'll do my best...

X is a discrete random variable, something that can take different values, like the number on a dice

P(X=a) means the probability that X takes a certain value a; for example on a dice P(X=5) is the probability that you roll a 5, which is 1/6

E(X) is an average value for X; if you rolled the dice loads of times and worked out the mean score, E(X) is what you'd expect it to be

So E(X) is just like another value for X

In the problem you posted above, the only values that X can take are 4, 6, 7, 10 and 11

E(X) is 7.4: X obviously can't be exactly 7.4, but it can be some values bigger than 7.4. The only values it can take bigger than 7.4 are 10 and 11

So the probability that X>E(X) = the probability that takes a value bigger than E(X), and the only values of X for which this is possible are 10 and 11

so P(X>E(X)) = P(X = 10 or 11) = P(X = 10) + P(X=11)

From the probability table you posted, P(X=10) = 0.125 and P(X=11) = 0.2, which is where that came from

Sorry for the long-winded post, but maybe you understand a bit better now?

Sorry for the time delay too - I was eating tea!
Thanks that makes sense now,

One thing - how do you know whether to multiply P(X=10) and P(X=1!) or to add them? I realise the difference is AND and OR, but not sure which one it is asking for in this question?

And.. seriously, thanks for all your help!
19. (Original post by Darkest Knight)
Thanks that makes sense now,

One thing - how do you know whether to multiply P(X=10) and P(X=1!) or to add them? I realise the difference is AND and OR, but not sure which one it is asking for in this question?

And.. seriously, thanks for all your help!
It is asking for OR. X>E(X) means that X=10 or X=11. X doesn't equal 10 and 11 - that would be impossible as you are only doing the test once.

For example, if rolling a dice, the probability that you get more than 4 on one throw equals the probability that you get 5 OR 6, so you add the probabilities. However, if you rolled a dice twice, the probability that you get a six both times equals the probability that you get a six on the first throw AND a six on the second throw, so you multiply the probabilities.

Don't worry about the help - its a reasonable excuse to have a break from chemistry coursework

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