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There are four people in an elevator, four floors in the building, and each person exits at random. Find the probability that:

a) all exit at different floors

b) all exit at the same floor

c) two get off at one floor and two get off at another

For a) I found $4!$ ways for the passengers to get off at different floors, so $$\frac{4!}{4^4} \text{would be the probability} = \frac{3}{32}$$

For b) there are only four ways for them to all exit on the same floor, so $$\frac{4}{256} = \frac{1}{64}$$

For c) am I allowed to group the $4$ people so that I am solving for $2$ people technically? For two people there would be $12$ possibilities, and there are three ways to group the $4$ individuals, so $$\frac{12 \cdot 3}{256} = \frac{9}{64}$$

I'm not sure if I'm doing these right, can you please check? Thank you.

RandomUser
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user148748
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    For the third, I would probably argue thus: the two floors can be chosen in $\binom{4}{2}$ ways. For each way, the $2$ people who get off at the lower of these floors can be chosen in $\binom{4}{2}$ ways, for a total of $\binom{4}{2}\binom{4}{2}$ (same result as yours). – André Nicolas May 08 '14 at 16:51

4 Answers4

3

Let us approach part (c) constructively. You appear to have correct answers for all the parts, but perhaps this will help.

First we select two of the four floors, which can obviously be done in $\dbinom{4}{2} = 6$ ways. Now we select two people from the four people to leave on the first floor, which can be done in $\dbinom{4}{2} = 6$ ways. We see that there are a total of $6 \times 6 = 36$ successful outcomes.

There are a total of $4^{4} = 256$ ways. Our probability is thus $\frac{36}{256} = \boxed{\frac{9}{64}}.$

K. Jiang
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Your calculation for $c)$ is correct, though, as the comments and answers show, you seem not to have expressed it in a readily comprehensible manner.

I would express what I understand your argument to be as: There are $3$ ways to split the $4$ people into $2$ groups. Then we can treat those two groups like $2$ people and choose the floors they get off at, $4$ options for the first group and $3$ for the second, or $12$ overall, for a total of $3\cdot12=36$ options out of $4^4=256$.

joriki
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Solutions using conditional probabilities:

Let's enumerate people $P_1, P_2, P_3, P_4$.

a) all exit at different floors

$P_1$ can pick any of 4 floors with probability $\frac{1}{4}$ each, so we have 4 cases. Because all 4 cases are equivalent we focus on just one. Now, given the floor $P_1$ exited at, $P_2$ can exit on any of the 3 floors left with probability $\frac{3}{4}$, then $P_3$ can exit on any of the 2 floors left with probability $\frac{2}{4}$, and finally, $P_4$ only one floor left with probability $\frac{1}{4}$. Thus, multiplying these conditional probabilities together:

$\sum_{floor=1,4}{\frac{1}{4} * \frac{3}{4} * \frac{2}{4}} * \frac{1}{4} = \frac{3}{4} * \frac{2}{4} * \frac{1}{4} = \frac{6}{64} = \boxed{\frac{3}{32}}$

b) all exit at the same floor

Again, $P_1$ can pick any of 4 floors with probability $\frac{1}{4}$ each, and then the rest of people have a probability $\frac{1}{4}$ to match that floor:

$\sum_{floor=1,4}{\frac{1}{4} * \frac{1}{4} * \frac{1}{4} * \frac{1}{4}} = (\frac{1}{4})^3 = = \boxed{\frac{1}{64}}$

c) two get off at one floor and two get off at another

Given $P_1$ picked a floor, we have 4 cases with probability $\frac{1}{4}$ each: $P_2$ picks the same floor with probability $\frac{1}{4}$, and the other two a different floor with probability ($\frac{3}{4})^2$:

$\sum_{floor=1,4}{\frac{1}{4} * (\frac{3}{4})^2} = \frac{1}{4} * (\frac{3}{4})^2 = = \boxed{\frac{9}{64}}$

Now, logical question would be what if it's not $P_2$ that exits on the same floor with $P_1$ but rather $P_3$ or $P_4$? There are $3!$ combinations that may happen with equal probabilities of $\frac{1}{3!}$, which would eliminate each other. But, to be complete here is the full formula:

$\sum_{floor=1,4}{\frac{1}{4} * \sum_{cases=1,3!}{\frac{1}{3!} * \frac{1}{4} * (\frac{3}{4})^2}} = \frac{1}{4} * (\frac{3}{4})^2 = = \boxed{\frac{9}{64}}$

The same reduction, of course, applied to a) and b).

topchef
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a and b are correct. For c, I would say there are ${4 \choose 2}$ ways to choose the first pair, $4$ ways to choose the floor they get off on, and $3$ ways to choose the floor the other pair gets off on, but we need to divide by $2$ because we can swap the pairs. That gives $\frac {6\cdot 4 \cdot 3}{2 \cdot 4^4}=\frac 9{64}$ This is the same as your answer, but I don't understand the logic that gives $12$ possibilities for a pair. We don't care about the order of selection in a pair.

Ross Millikan
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