This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ....
View
Full
Document
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).
 Fall '10
 Beaulieu
 Math, Probability

Click to edit the document details