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
x
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Darkest Knight
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 15112008 17:19

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 15112008 17:34
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 thenLast edited by sonofdot; 15112008 at 17:48. 
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 15112008 17:42
(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
is that okay? 
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 15112008 17:47
(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?
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 
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 15112008 17:49
I don't understand how your working out the table. I thought you multiply by the number and the top by number at left...

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 15112008 17:52
(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...
Sorry if that just confused you more! 
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 15112008 18:06
(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!
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 
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 15112008 18:14
(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
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? 
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 15112008 18:17
(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? 
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 15112008 18:50
Sorry one last thing.. (and +rep given by the way)
How can I find the probability that X > E(X)?
Thanks! 
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 15112008 19:12
(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!
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)... 
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 15112008 19:15
(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)...
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 
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 15112008 19:17
(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 
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 15112008 19:21
(Original post by sonofdot)
Then P(X > E(X)) = P(X=10) + P(X=11) 
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 15112008 19:23
(Original post by Darkest Knight)
The books says the answer is 0.325!
P(X=11) = 0.2
P(X>E(X)) = P(X=10) + P(X=11) = 0.125 + 0.2 = 0.325 
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 15112008 19:24
(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
How does P(X=10) = 0.125
Bit confused where the "X =" comes from too..
Sorry! 
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 15112008 20:38
(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!
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 longwinded post, but maybe you understand a bit better now?
Sorry for the time delay too  I was eating tea! 
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 15112008 22:25
(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 longwinded post, but maybe you understand a bit better now?
Sorry for the time delay too  I was eating tea!
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! 
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 15112008 22:45
(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!
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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