hw3_print

hw3_print - STAT416 Probability — Spring 2011 Assignments...

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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 doesn’t work, B =nothing interesting on TV, and C =she doesn’t 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) . 3B–4G–5R. 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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hw3_print - STAT416 Probability — Spring 2011 Assignments...

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