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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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39% (02:15) correct
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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
Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are rando
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18 Jul 2022, 08:48
18 Jul 2022, 08:48
6
Number of ways to distribute hat = 4!
Number of ways atleast one gets a hat = One gets + Ways 2 get + ways 3 get + way 4 get
= 4C1 + 4C2 + 4C3 + 4C4
Probability that no one gets = 1- probability that at least one gets
= 1 - (4C1 + 4C2 + 4C3 + 4C4)/4!
= 9/24 = 3/8
Number of ways atleast one gets a hat = One gets + Ways 2 get + ways 3 get + way 4 get
= 4C1 + 4C2 + 4C3 + 4C4
Probability that no one gets = 1- probability that at least one gets
= 1 - (4C1 + 4C2 + 4C3 + 4C4)/4!
= 9/24 = 3/8
General Discussion
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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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
