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Unformatted text preview: STAT416: Probability Spring 2011 Assignments 1 Homework #3 (Solution) February 8, 2011 1 Total possible points is 45 1. Ex. 3.2 (5 pts) . P ( Sat ) = 0 . 6, P ( Sun ) = 0 . 8, P ( Sat c Sun c ) = 0 . 1. P ( Sun  Sat ) = P ( Sun Sat ) P ( Sat ) . But P ( Sat c Sun c ) = P ([ Sun Sat ] c ) = 1 P ( Sun Sat ) = 1 [ P ( Sun ) + P ( Sat ) P ( Sun Sat )] P ( Sun Sat ) = 0 . 6 + 0 . 8 . 9 = 0 . 5. Therefore, P ( Sun  Sat ) = P ( Sun Sat ) P ( Sat ) = . 5 . 6 = 5 6 = 0 . 833 . 2. Ex. 3.6 (5 pts) . Define the events A =computer doesnt work, B =nothing interesting on TV, and C =she doesnt pick up the phone. Then, P (Sam goes to bed at 8pm) = P ( A B C ) = P ( C  A B ) P ( B  A ) P ( A ) = 0 . 1 . 3 . 75 = 0 . 0225 . 3. Ex. 3.8 (5 pts) . 3B4G5R. There is a total of 12 balls. P (both red  same color) = P ( RR ) P ( RR ) + P ( BB ) + P ( GG ) = 5 12 4 11 5 4 12 11 + 4 3 12 11 + 3 2 12 11 = 10 19...
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 Spring '08
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 Probability

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