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Al, Bob, Cal and Don each own 1 hat. If the 4 hats are rando

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Al, Bob, Cal and Don each own 1 hat. If the 4 hats are rando [#permalink] New post 07 Nov 2018, 07:15
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20% (00:01) correct 80% (01:23) wrong based on 5 sessions
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
[Reveal] Spoiler: OA

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2 KUDOS received
GRE Instructor
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Joined: 10 Apr 2015
Posts: 1177
Followers: 44

Kudos [?]: 1053 [2] , given: 6

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Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are rando [#permalink] New post 08 Nov 2018, 06:01
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GreenlightTestPrep 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


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
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Brent Hanneson – Creator of greenlighttestprep.com
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Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are rando   [#permalink] 08 Nov 2018, 06:01
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Al, Bob, Cal and Don each own 1 hat. If the 4 hats are rando

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