HW24Answers(8.3) - Math331 Spring 2008 Instructor David...

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Unformatted text preview: Math331, Spring 2008 Instructor: David Anderson Section 8.3 Homework Answers Homework: pgs. 353 - 355, #’s 1, 3, 5 (only discrete case), 13. 1. We have that p ( x,y ) = braceleftbigg 1 25 ( x 2 + y 2 ) if x = 1 , 2 , y = 0 , 1 , 2 else . We want, p X | Y ( x | y ) = p ( x,y ) p Y ( y ) . So we need to find p Y ( y ) for y ∈ { , 1 , 2 } . p Y ( y ) = 2 summationdisplay x =1 1 25 ( x 2 + y 2 ) = 1 25 (5 + 2 y 2 ) . Thus, for x ∈ { 1 , 2 } , y ∈ { , 1 , 2 } , p X | Y ( x | y ) = 1 25 ( x 2 + y 2 ) * 25 5 + 2 y 2 = x 2 + y 2 5 + 2 y 2 . To find P ( X = 2 | Y = 1) is now simple: P ( X = 2 | Y = 1) = p X | Y (2 , 1) = 2 2 + 1 2 5 + 2 * 1 2 = 5 7 . Similarly, the expectation is given via E [ X | Y = 1] = 2 summationdisplay x =1 xp X | Y ( x | 1) = 1 * 1 + 1 7 + 2 * 5 7 = 12 7 . 3. Let X be the number of flips needed until the 6th head appears. Let Z be the number of flips needed to get the third head. We know that Z = 5. Let Y be the number of flips after the fifth until the 3rd head appears. Then, because we know that the third head occurs onuntil the 3rd head appears....
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This note was uploaded on 06/24/2008 for the course MATH 331 taught by Professor Anderson during the Spring '08 term at University of Wisconsin.

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HW24Answers(8.3) - Math331 Spring 2008 Instructor David...

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