Discussion

Let the joint distribution of (X, Y) be specified by :

Y \ X	1	2	3	4
1	1/16	3/32	1/8	5/32
2	3/32	1/8	5/32	3/16

Then \(P[X = 2Y]\) is :

Answer: B

\(P[X = 2Y] = \) 9/32.

\(X = 2Y\) happens only at \((X, Y) = (2, 1)\) and \((4, 2)\):

\[ \begin{aligned} \frac{3}{32} + \frac{3}{16} &= \frac{3}{32} + \frac{6}{32} \\ &= \frac{9}{32} \end{aligned} \]

(The table's total is 1, as it should be.) 0 ignores these cells, and 11/32 or 1/2 add other cells, such as (2, 2) or (4, 1), where \(X \ne 2Y\).

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