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Challenge: Al, Bob, Cal and Don each own 1 hat. If the 4...

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Challenge: Al, Bob, Cal and Don each own 1 hat. If the 4...

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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

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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

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Brent Hanneson – Creator of GMATPrepNow.com
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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) $$=\frac{9}{24}=\frac{3}{8}$$
$$answer\ is\ option\ D$$

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