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#1
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I'm trying to solve this problem by calculating the probability of each outcome, and finding the expected value based on that, but I'm messing up my probabilities.
I get the possible values as 2, 4, 5, 7, and 10. I get their respective probabilities as (1/3), (1/9), (1/9), (5/18), and (1/9). I realize these don't sum to 1, and that's my problem...I've double checked everything and not found the mistake. Does anyone know what the correct probabilities should be? |
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#3
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It looks like he is trying to calculate the prior probabilities, which should sum to 1.
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#8
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Pr(X=2) = 0.5(2/3^2 + 1/3*2/3) = 1/3
Pr(X=4) = 0.5(1/3*2/3*1/3 + 2/3*2/3*1/3) = 1/9 Pr(X=5) = 0.5(2/3*1/3*2) = 2/9 Pr(X=7) = 0.5(1/3*(2/3^2 + 1/3^2) + 2/3(2/3^2 + 1/3^2)) = 5/18 Pr(X=10) = 0.5(1/3*(1/3*2/3) + 2/3*(1/3*2/3)) = 1/9 ok, where is my math mistake? This is why doing the problem like this is dumb! |
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#9
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Quote:
4: 1/2 * 1/3 * 2/9 + 1/2 * 2/3 * 2/9 = 1/9 5: 1/2 * 2/3 * 1/3 + 1/2 * 1/3 * 1/3 = 1/6 7: 1/2 * 1/3 * 5/9 + 1/2 * 2/3 * 5/9 = 5/18 10: 1/2 * 1/3 * 1/3 * 2/3 + 1/2 * 2/3 * 1/3 * 2/3 = 1/9 1/3 + 1/9 + 1/6 + 5/18 + 1/9 = 1 |
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#10
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found it, thanks pcact.
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