fall 2005 STAT 230 Test2

fall 2005 STAT 230 Test2 - 1(a) PX 0|A . 75 3 . 421 88, by...

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1(a) P X 0| A .75 3 . 42188,by independence (b)We want to calculate P A | X 0 . Note that: P A | X 0 P X 0| A P A P X 0 , by Bayes rule P X 0| A was calculated in part (a). P A 0.4, given information To calculate P X 0 use the partion rule P X 0 P X 0| A P A P X 0| B P B P X 0| C P C .75 3 0.4 .8 3 0.2 .85 3 0.4 (c) X can take on the values: 0,1,2,3 To find f x let us use the partition rule once again, f x P X x P X x | A P A P X x | B P B P X x | C P C 3 x .25 x .75 3 x 0.4 3 x .2 x .8 3 x 0.2 3 x .15 x .85 3 x 0.4 for x 0,1,2,3. (d)
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fall 2005 STAT 230 Test2 - 1(a) PX 0|A . 75 3 . 421 88, by...

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