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# squiz1 - teams we get the same game so the total number of...

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Solutions to quiz # 1 (January 12) 1. a) We can choose a committee of 3 from a group of 10 people in ± 10 3 ² = 10! 3!7! ways and then we can choose a committee of 4 from the group of remaining 7 people in ± 7 4 ² = 7! 4!3! ways. Altogether, there are 10! 3!7! · 7! 4!3! = 10! 4!3!3! = 4200 ways to choose the committees. Answer: the two committees can be chosen in 10! 4!3!3! ways. b) We can choose the ﬁrst team in ± 10 4 ² = 10! 4!6! ways, after which we can choose the second team in ± 6 4 ² = 6! 4!2! ways. Hence the total number of ways to choose ﬁrst the ﬁrst team and then the second team is 10! 4!6! · 6! 4!2! = 10! 4!4!2! ways. However, the order in which the teams are chosen does not matter (if we switch the
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Unformatted text preview: teams we get the same game), so the total number of diļ¬erent games is 1 2 Ā· 10! 4!4!2! = 1575 . Answer: there are 10! 4!4!2!2! diļ¬erent games. 2. The sample space is { AB,AC,BC } . We have P ( { AB,AC } ) = P ( AB ) + P ( AC ) = 5 8 , P ( { AB,BC } ) = P ( AB ) + P ( BC ) = 5 8 and P ( AB ) + P ( AC ) + P ( BC ) =1 . Therefore, P ( { AC,BC } ) = P ( AC ) + P ( BC ) =2 Ā³ P ( AB ) + P ( AC ) + P ( BC ) Ā“-Ā³ P ( AB ) + P ( AC ) Ā“-Ā³ P ( AB ) + P ( BC ) Ā“ =2-5 8-5 8 = 3 4 . Answer: The probability that one of the chosen courses is Communications is 3 / 4. 1...
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