# How to find the pmf of the number of students who get their passport back?

$$4$$ passports are returned randomly to $$4$$ students. Let the random variable $$X$$ be the number of students who get their passport back. Determine the pmf of $$X$$.

$$p(0) = \frac{9}{24}, p(1) = \frac{1}{3}, p(2) = \frac{1}{4}, p(3) = 0, p(4) = \frac{1}{24}$$

I understand that $$p(3) = 0$$ because if $$3$$ people get the correct passport back then that implies the remaining person got the wrong passport which is not possible since the remaining passport must be the correct passport. I think my answer makes more sense even though it is incorrect. My question is how is the answer above obtained and why is my answer incorrect?

Let $$e=$${at least one person gets their passport back}$$=$${$$e_1$$,$$e_2$$,$$e_3$$,$$e_4$$} where $$e_i$$ is the subevent that $$i$$th person gets back their passport.

So the pmf $$P(X=e)=p(e)$$ is

$$p(0) = p(\bar e)=1-p(e)=1-[p(1)+p(2)+p(3)+p(4)]$$ $$p(1) = \frac{\binom{4}{1}}{4!}, p(2) = \frac{\binom{4}{2}}{4!}, p(3) = \frac{\binom{4}{3}}{4!}, p(4) = \frac{\binom{4}{4}}{4!}$$

which simplifies to

$$p(0) =\frac{9}{24}, p(1) = \frac{4}{24}, p(2) = \frac{6}{24}, p(3) = \frac{4}{24}, p(4) = \frac{1}{24}$$

Only $$p(1)$$ and $$p(3)$$ are incorrect. Although the correct answer for $$p(3)=0$$, I think $$p(3)=\frac{4}{24}$$ because technically $$4$$ people are getting back their passports if $$3$$ people get back the correct passport. Can someone explain why my answer is wrong? Also my probabilities still add up to $$1$$ although it is incorrect.

$$p(3) = 0$$ because as you mentioned, if three people get their passport back then the remaining person cannot get the wrong passport, hence this event never occurs.

Your answer for $$p(1)$$ is close to the correct answer. You have to multiply $$p(1)=\frac{\binom{4}{1}}{24}$$ by 2 because $$\binom{4}{1}$$ is the number of ways to select one person to get the correct passport out of four possible people. This is equivalent to the number of people who get the incorrect passport. For instance, let the three remaining people/passports be $$p_1,p_2,p_3$$. There are 2*$$\binom{4}{1}$$ ways they can get the incorrect passport.

For example,

• if $$p_1$$ gets $$p_2$$'s passport then
• $$p_2$$ gets $$p_3$$'s passport and
• $$p_3$$ gets $$p_1$$'s passport.

This can happen in $$2$$ ways, hence $$p(1)=\frac{2*\binom{4}{1}}{24}=\frac{8}{24}$$

Let me show you a very simple way to solve the problem. You have to calculate 5 probabilities

• $$P(0)$$

• $$P(1)$$

• $$P(2)$$

• $$P(3)$$

• $$P(4)$$

$$P(3)=0$$, trivial because, as you noted, if 3 passports are correct it is not possible to have the fourth wrong.

$$P(0)=\frac{9}{24}$$ as there are

$$!4=4!\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}\right)=9$$ favourable ways to have permutations of elements all wrong on 24 possible ways

$$P(4)$$, conversely, there is only 1 favourable permutation on the 24 possible

with similar reasoning as $$P(0)$$

$$P(1)=\frac{ !3}{6}=\frac{1}{3}$$

$$P(2)$$ can be derived as complement probability to 1

• Your formula is $$\frac{combinations}{permutations}$$. As you know $$combinations=\frac{permutations}{redundancies}$$, so each probability that you calculated is $$\frac{combinations}{permutations}=\frac 1 {redundancies}$$ and doesn't have much meaning. For example, for $$\frac{4\choose 2}{4!}=\frac{1}{2!2!}$$ is the probability of picking a unique sequence when you've got two groups of two things, e.g. if we were to put A, B where the underscores are and C, D in the exclamation marks, _ ! ! _, you have calculated the probability of getting one unique sequence such as ACDB. The immediate relation to the problem is not clear.

• Instead, you should reason as follows. For p(1), there is 1/4 chance that the correct letter goes to student-1. Then the remaining letters have a 2/3 * 1/2 * 1 chance of going to an incorrect student. Now this can also happen to students, 2, 3, and 4, so multiply this probability by 4 to get $$\frac 1 4\left(\frac 2 3 *\frac 1 2* 1\right)*4=\frac 1 3$$. For p(3), it is not possible to get exactly 3 right matches because of exactly what you said. p(4) is a separate case than p(3). p(4) is when all 4 people get their passports back, but it so happens that it is not p(3) because p(3) is when 3 people get their passports back and 1 does not - there is an important distinction.