jacksonmeg
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Three independent values of X are chose, X1, X2, X3, it is given that X1+X2+X3 = 19, find sets of values for X1, X2 and X3 and find the probability P(X=7)

x 1 3 5 7
P(x=x) 0.4 0.3 0.2 0.1

I have the set of values as 775, 757, 577, now what do I do? :/
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ghostwalker
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(Original post by jacksonmeg)
Three independent values of X are chose, X1, X2, X3, it is given that X1+X2+X3 = 19, find sets of values for X1, X2 and X3 and find the probability P(X=7)

x 1 3 5 7
P(x=x) 0.4 0.3 0.2 0.1

I have the set of values as 775, 757, 577, now what do I do? :/
Do you have a link to the original question, or its exact wording. To find the probability X=7, doesn't make sense. P(X1=7) given the previous, for instance, would be a reasonable question to ask.
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jacksonmeg
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(Original post by ghostwalker)
Do you have a link to the original question, or its exact wording. To find the probability X=7, doesn't make sense. P(X1=7) given the previous, for instance, would be a reasonable question to ask.
its june 2013 and I can't link it, i'll type it out exactly

Three independent values of X, denoted by X1, X2, X3 are chosen
Given X1 + X2 + X3 = 19, write down al possible sets of values for X1, X2, X3 and hence find P(X1=7)
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ghostwalker
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(Original post by jacksonmeg)
its june 2013 and I can't link it, i'll type it out exactly

Three independent values of X, denoted by X1, X2, X3 are chosen
Given X1 + X2 + X3 = 19, write down al possible sets of values for X1, X2, X3 and hence find P(X1=7)
OK. All three of those possibilities (775, 757, 577) are equally likely. Two of them have X1=7, hence the probability X1=7, given those three, is....
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Lamalam
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(Original post by ghostwalker)
OK. All three of those possibilities (775, 757, 577) are equally likely. Two of them have X1=7, hence the probability X1=7, given those three, is....
Sorry for hijacking the thread. Is it just 2/3?

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ghostwalker
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(Original post by Lamalam)
Sorry for hijacking the thread. Is it just 2/3?

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Yep.
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