Math121 Discussion - 9th week solutions

# Math121 Discussion - 9th week solutions - MATH 121...

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Unformatted text preview: MATH 121 DISCUSSION WEEK 9 SOLUTIONS 1) (a) S = n { Al, Bo } , { Al, Dee } , { Al, Fay } , { Bo, Dee } , { Bo, Fay } , { Dee, Fay } o (b) A = n { Al, Dee } , { Bo, Dee } , { Dee, Fay } o (c) P(A) = 3 6 = 1 2 . 2) (a) S = { abc, acb, bac, bca, cab, cba } (b) A = { abc, acb, bac } (c) The event that b and c are next to each other is: B = { abc, acb, bca, cba } . Thus P(B) = 4 6 = 2 3 . 3) P(at least two share a birthday) = 1. This is a certain event. (Even if there were 365 people in the class who have different birthdays, the class has 367 students, which means that the 366th student must have a birthday in common with at least one of the 365 students considered before him/her. So it is always the case that at least two people have a common birthday.) 4) P(draw 5 aces) = 0. This is an impossible event since a fair deck only has 4 aces. 5) (a) P(draw clubs, diamonds, 6 or a Queen) = P(draw clubs) + P(draw diamonds) + P(draw a 6) + P(draw a Queen) - P(repeats). The repeated elements are { 6 of clubs, 6 of diamonds, Queen of clubs, Queen of diamonds } , and the probability of drawing any of these repeated cards is 4 52 . Thus, we have: P(draw clubs, diamonds, 6 or a Queen) = ( 13 52 + 13 52 + 4 52 + 4 52 )- 4 52 = 30 52 = 15 26 . (b) P(draw Kings first time ∩ draw Kings second time ∩ draw Kings third time) = P(draw Kings first time) × P(draw Kings second time) × P(draw Kings third time) = ( 4 52 )( 4 52 )( 4 52 ) = ( 4 52 ) 3 = 1 2197 ....
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## This note was uploaded on 12/07/2010 for the course MATH Math 121 taught by Professor Beaulieu during the Fall '10 term at UMass (Amherst).

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Math121 Discussion - 9th week solutions - MATH 121...

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