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Statistical Techniques for Transportation Engineering by Naresh Davergave, Anil Shah, G Rao, Kumar Molugaram

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2. F(1,3)=P(0,1)+P(0,2)+P(0,3)+P(1,1)+P(1,2)+P(1,3)=124+112+112=112+16+16=1524=58image

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

1. P(X≤1)

2. P(Y≤3)

X\Y123456
000132image232image232image332
111611618181818
21321321641640264

Solution:

1. P(X1)=

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