Thread: Find the distribution
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Old November 2nd, 2009, 03:39 PM
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Quote:
Originally Posted by craig View Post
Y = X_1 + X_2, therefore Y can take the values 0..4.

P(Y=0) = \frac{1}{8}\times\frac{1}{8} = \frac{1}{64}

P(Y=1) = 2(\frac{3}{8}\times\frac{1}{8}) = \frac{3}{32}

P(Y=2) = \frac{17}{64}

P(Y=3) = \frac{3}{8}

P(Y=4) = \frac{1}{4}

Is this what you meant to do?

Thanks for the reply
Yes.
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