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 hatLet 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, Db, 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 outcomesWe 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
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?