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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are rando
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07 Nov 2018, 07:15
07 Nov 2018, 07:15
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Question Stats:
38% (02:17) correct
61% (02:15) wrong based on 78 sessions
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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
A) 1/8
B) 1/4
C) 1/3
D) 3/8
E) 1/2
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are rando
[#permalink]
08 Nov 2018, 06:01
08 Nov 2018, 06:01
5
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
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 pretty fast.
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
At this point, we can see that there are 3 outcomes when Al receives Bob's hat. So, we can conclude that there will also be 3 outcomes when Al receives Cal's hat, and 3 outcomes when Al receives Don's hat
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
General Discussion
Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are rando
[#permalink]
18 Mar 2022, 23:14
18 Mar 2022, 23:14
GreenlightTestPrep wrote:
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
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
Hi, Brent!
Could you explain mathematically why 3/4*2/3*1/2 doesn't work to find the probability in this case? I can see that your answer is correct, it is because you have laid out all the possible scenarios that satisfy the question.
What would be the approach if there were 13 hats and the probability to distribute all of them incorrectly to their owners?
