Four friends, Adin, Brigid, Chioma, and Draven, go to a restaurant and each order one dish. The waiter took down the order but forgot to mark for whom each dish was for, and hence served a random dish to each of the four patrons. What is the probability that none of the customers got the dish they ordered?
Enter your answer as a fraction reduced to the simplest terms.
Here total number of ways waiter can distribute the dish is 4! = 24
then there are 3 ways the distribution can be messed up so 3*4 = 12
and hence
happy cases
Total cases
12/24 = 1/2 is I am counting happy cases wrong?
No. of total cases = 4! = 24
No. of favorable cases = 3 + 3 + 3 = 9 (when B is in first slot, when C is in first slot, when D is in 1st slot)
You can just brute force for one scenario -
So consider - (B) (A) (C) (D) assuming A gets B’s dish, B gets A’s dish, C and D get their own dish. This case fails since C and D get their own dish, so try other 5 scenarios keeping B in first slot.
We know this scenario can happen in 6 ways(i.e., A gets B’s dish, so other 3 dishes can happen in 6 different ways):
BACD - fails (already tried above)
BADC - works
BCAD - fails (since D gets his dish)
BCDA - works
BDAC - works
BDCA - fails (C gets his dish)
So 3 cases work for B.
Same thing applies when C(3 more cases) and D(3 more cases) are in first slot as well (so total happy outcomes will be 3+3+3 = 9).
Don’t forget we cannot have A in the first slot because we don’t require A getting served with A’s dish. So we can just eliminate all the 6 cases when A is in the first slot.
(# of favorable outcomes) / (total no. of outcomes) = 9/24 = 3/8
Hi guys,
I was doing some questions from Greg’s new Quant list. Came across the following
Four friends, Adin, Brigid, Chioma, and Draven, go to a restaurant and each order one dish. The waiter took down the order but forgot to mark for whom each dish was for, and hence served a random dish to each of the four patrons. What is the probability that none of the customers got the dish they ordered?
It is a Numeric Entry question. My answer was 81/256.
I arrived at the answer in 2 ways
Approach 1:
For each individual, P(Getting Correct Dish) = 1/4 and P(Getting wrong dish) = 3/4
So P(none get the correct dish) = \frac{3}{4} \times \frac{3}{4} \times \frac{3}{4} \times \frac{3}{4}
Approach 2:
P(none get correct dish) = 1 - P(Atleast 1 gets correct dish)
1 - (4C0 + 4C1 + 4C2 + 4C3 + 4C4)* (1/256) = 1 - (175/256) = 81/256
Can anyone help me with where I’m going wrong here?
Your answer assumes that each person has a choice of 4 dishes, one of which is their “chosen dish”. The problem is that they are non-repeating, that is, A and B cannot have the same dish.
The question says that each of them orders a dish, and the waiter took down the details of the 4 dishes. So when he was giving back the dishes to the patrons randomly, once he gives a dish to someone, he has one less dish left. You missed that part.
Yes. If A orders a dish, B should order another one. There is a slight ambiguity in the wording which should be fixed soon, but that does not affect correctness.
