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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## Challenge: Al, Bob, Cal and Don each own 1 hat. If the 4... tagged by: Brent@GMATPrepNow ##### This topic has 1 expert reply and 1 member reply ### GMAT/MBA Expert ## Challenge: Al, Bob, Cal and Don each own 1 hat. If the 4... ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult Al, Bob, Cal and Don each own 1 hat. If the 4 hats are randomly distributed so that each man receives exactly 1 hat, what is the probability that no one receives his own hat? A) 1/8 B) 1/4 C) 1/3 D) 3/8 E) 1/2 Answer: D Difficulty level: 650 - 700 Source: www.gmatprepnow.com _________________ Brent Hanneson â€“ Creator of GMATPrepNow.com Use my video course along with Sign up for free Question of the Day emails And check out all of these free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months! ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 12906 messages Followed by: 1248 members Upvotes: 5254 GMAT Score: 770 Brent@GMATPrepNow wrote: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are randomly distributed so that each man receives exactly 1 hat, what is the probability that no one receives his own hat? A) 1/8 B) 1/4 C) 1/3 D) 3/8 E) 1/2 Answer: D Difficulty level: 650 - 700 Source: www.gmatprepnow.com I created this question to highlight many students' tendency to avoid listing and counting as a possible approach. As you'll see, the approach is probably the fastest approach. P(no one receives his own hat) = (# of outcomes in which no one receives his own hat)/(TOTAL number of outcomes) # of outcomes in which no one receives his own hat Let a, b, c and d represent the hats owned by Al (A), Bob (B), Cal (C) and Don (D) Let's systematically list the HATS to be paired up with A, B, C, and D A, B, C, D b, a, d, c b, c, d, a b, d, a, c c, a, d, b c, d, a, b c, d, b, a d, a, b, c d, c, a, b d, c, b, a So, there are 9 outcomes in which one receives his own hat TOTAL number of outcomes We can arrange n unique objects in n! ways So, we can arrange the 4 hats in 4! ways (= 24 ways) So, there are 24 possible outcomes P(no one receives his own hat) = 9/24 = 3/8 Answer: D Cheers, Brent _________________ Brent Hanneson â€“ Creator of GMATPrepNow.com Use my video course along with Sign up for free Question of the Day emails And check out all of these free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months! ### Top Member Legendary Member Joined 02 Mar 2018 Posted: 1039 messages Followed by: 2 members Probability (no one receives his own hat) = numbers of outcomes in which no one receives his own hat ; Let Al's hat = a , Bob's hat = b, Cal's hat = c, Don = d For Al = b, c, d Bob = a, c, d Cal = a, b, c The 4 of them can pick up 3 different hats. 3 time without receiving their own hat hence outcome for not receiving own hat = 9 Outcome (3 hats * 3 times) Given that we can arrange the 4 hats in 4 different ways ; a, b, c, d b, c, d, a c, d, a, b d, a, c, b Total number of outcomes = 4 (!) ways $$=4!=4\cdot3\cdot2\cdot1=24ways\ or\ possible\ outcomes$$ Probability (no ones receives his own hat) $$=\frac{9}{24}=\frac{3}{8}$$ $$answer\ is\ option\ D$$ • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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