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# S1 help on discrete random variables! Watch

1. I cant seem to do this question, can somebody give me a hand? I can't seem to do part e!

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2. Basically you find all the combinations of what X1 and X2 could be that satisfy the inequality (i.e. they have to be more than 1, but equal to 3 or less), then you add up the probabilites of the combinations occurring:

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Combinations of (X1,X2) are (0,3) (0,2) (1,1) (1,2) (2,1) (3,0) & (2,0)

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k = 0.25, probabilities of 0 = 0.5, 1=0.25, 2= 0, 3 =0.25... so use the AND rule for probabilities to the combinations above
3. (Original post by Piguy)
Basically you find all the combinations of what X1 and X2 could be that satisfy the inequality (i.e. they have to be more than 1, but equal to 3 or less), then you add up the probabilites of the combinations occurring:

Spoiler:
Show
Combinations of (X1,X2) are (0,3) (0,2) (1,1) (1,2) (2,1) (3,0) & (2,0)

Spoiler:
Show
k = 0.25, probabilities of 0 = 0.5, 1=0.25, 2= 0, 3 =0.25... so use the AND rule for probabilities to the combinations above
I'm pretty sure there's no "inequality" as such for , combinations like (3,2) and (1,3) etc should still be possible.
4. (Original post by justinawe)
I'm pretty sure there's no "inequality" as such for , combinations like (3,2) and (1,3) etc should still be possible.
Woops, looking at f, silly me
5. In that case you just need to find the combinations of what X1 + X2 could be:

Spoiler:
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It's best to be logical about it:

0 = (0,0)

1 = (0,1) + (1,0)

2 = (2,0) + (0,2) + (1,1)

3 = (0,3) + (3,0) + (1,2) + (2,1)

4 = (1,3) + (3,1) + (2,2)

5 = (2,3) + (3,2)

6 = (3,3)

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Updated: May 15, 2013
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