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2. $\begin{array}{ll}F\left(1,3\right)\hfill & =P\left(0,1\right)+P\left(0,2\right)+P\left(0,3\right)+P\left(1,1\right)+P\left(1,2\right)+P\left(1,3\right)\hfill \\ =\frac{1}{24}+\frac{1}{12}+\frac{1}{12}=\frac{1}{12}+\frac{1}{6}+\frac{1}{6}=\frac{15}{24}=\frac{5}{8}\hfill \end{array}$

Example 4.17: From the following table for bivariate distribution. Find

1. P(X≤1)

2. P(Y≤3)

 X\Y 1 2 3 4 5 6 0 0 0 $\frac{1}{32}$ $\frac{2}{32}$ $\frac{2}{32}$ $\frac{3}{32}$ 1 $\frac{1}{16}$ $\frac{1}{16}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ 2 $\frac{1}{32}$ $\frac{1}{32}$ $\frac{1}{64}$ $\frac{1}{64}$ 0 $\frac{2}{64}$

Solution:

1. $\begin{array}{ll}P\left(X\le 1\right)\hfill & =\hfill \end{array}$

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