$\DeclareMathOperator\Var{Var}\DeclareMathOperator\Bern{Bern}\DeclareMathOperator\Pois{Pois}$Question: What not-trivial mistakes do students often make when solving problems in probability theory, mathematical statistics and random processes?

Some examples of wrong solutions:

Problem 1: Find distribution of $F_{\xi}(\xi)$ for continous $F_{\xi}$.

"Solution": $F_{\xi}(\xi) = P(\xi \le \xi) = 1$.

Problem 2: Is it possible to guess if one of a pair of random numbers is larger with probability ${}> \frac{1}2$?

"Solution". Obviously no (sometimes with some blurry reasoning, mentioning symmetry). (if smb.is interested, see discussion, e.g., in https://mathoverflow.net/questions/9037/how-is-it-that-you-can-guess-if-one-of-a-pair-of-random-numbers-is-larger-with)

Problem 3: Find $\Var(\min(X,Y))$ for independent $X,Y \sim \exp(1)$.

"Solution". If $X \le Y$ then $\min(X,Y) = X$ and $\Var(\min(X,Y)) = DX = 1$, if $X > Y$ then $\min(X,Y) = Y$ and $\Var(\min(X,Y)) = DY = 1$, so in any case we got $\Var(\min(X,Y)) = 1$. A more absurd version is problem 3b: find $D\xi$ for $\xi \sim \Bern(p)$. "Solution": $\xi$ takes values $0$ and $1$, if it's equal to $0$, then $\Var\xi = \Var(0)=0$, in other case $\Var\xi = \Var(1)=0$, hence $\Var(\xi)=0$.

Problem 4: Suppose that systems of sets $\mathcal{A}_1$ and $\mathcal{A}_2$ are independent. Are $\sigma(\mathcal{A}_1)$ and $\sigma(\mathcal{A}_2)$ independent? "Solution". Obviously yes (sometimes with some blurry reasoning).

Problem 5: $2n$ teams were divided to $2$ subgroups with $n$ teams in each group. What is the chance that the $2$ strongest teams will play in the same subgroup?

"Solution": $\frac12$ by symmetry. (See discussion, e.g. Probability problem: what's the chance that two strongest teams will play in the same subgroup.)

Problem 6: "A patient goes to see a doctor. The doctor performs a test with 99 percent reliability—that is, 99 percent of people who are sick test positive and 99 percent of the healthy people test negative. The doctor knows that only 0.01 percent of the people in the country are sick. If the patient test is positive, what are the chances the patient is sick?"

"Solution": 99 percent as follows immediately from the task.

Some references: There are different paradoxes, such as Monty Hall problem, see, e.g. "Paradoxes in Probability Theory and Mathematical Statistics" by G. J. Székely. There are some interesting examples of popular mistakes in "The evolution with age of probabilistic, intuitively based misconceptions" by E. Fischbein and D. Schnarch (and references therein).

Of course, there are a huge number of mistakes that, in principle, can be made, but I mean, firstly, popular ones, and secondly, it is desirable that these were errors not on the most simple topics of combinatorics and not connected with typos and so on. What not-trivial mistakes did you see lots of times?

The question is connected with Common misconceptions in high school probability curriculum

Addition: uninteresting examples :

  1. A coin is tossed twice at random. What is the probability of getting the same face? "Solution:" Three possible outcomes are HH, HT=TH, TT, where H = head, T = tail. So the probability is $\frac13$.

  2. $E\xi^2 = (E \xi)^2$ because it's "obvious" because we are speaking about mean, and even reference: there was an equality $E \xi \eta = E\xi E \eta$ in one of the properties of expectation.

  3. For independent $\xi$ and $\eta$ we have $\Var(\xi + \eta) = \Var(\xi) + \Var(\eta)$ "hence" $\Var(\xi - \eta) = \Var(\xi) - \Var(\eta)$.

  4. Density of $U[0,1]$ is $F' = I_{[0,1]}(x)$ a.e., so density of $\Pois(\lambda)$ is $F'= 0$ a.e.

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    $\begingroup$ I did not downvote, but I am guessing these downvotes are due to the question being too broad. It would be nice if people explained their downvotes. I hope this question being poorly received doesn't discourage you from participating more in this community,! $\endgroup$ Commented Jan 25, 2023 at 18:47
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    $\begingroup$ @StevenGubkin Thank you! About downvotes - it's interesting that the same question in researchers community mathoverflow ( mathoverflow.net/questions/439066/… ) was treated differently and they said that it's better to write it at matheducators.stackexchange.com/ :) $\endgroup$ Commented Jan 25, 2023 at 19:01

2 Answers 2


These two concepts from elementary probability are not that elementary to digest.

  1. If, in a probability experiment, event B's outcomes are restricted due to knowing that event A happens, the two events can still be independent.

    $\quad$ Intuition on the independence of two events

    As an example, consider a roll of a fair die: $ Ω= \{1,2,3,4,5,6\}\\ A= \lbrace 3, 6\rbrace\\ B= \lbrace 2, 4, 6 \rbrace$

    Knowing that $A$ happens restricts the outcomes of $B$ to just $\{6\}.$

    $P(A\cap B)=P(A)P(B),$ so $A$ and $B$ are independent events.

  2. For uncountable sample spaces, mutually exclusive events need not be disjoint (unless the author defines mutual exclusivity using something like $X\cap Y=\emptyset$ instead of something like $P(X\cap Y)=0$).

    For example, the events $(1,2]$ and $[2,3)$ are mutually exclusive but not disjoint.

  3. Two mutually exclusive positive-probability events are necessarily dependent.

    It is common to naively make wrong inferences between independence and mutual exclusivity, so this illustration may appear counterintuitive: $$ \begin{array}{r} \begin{array}{c|c|c} \style{font-family:inherit}{} & \style{font-family:inherit}{U_1} & \style{font-family:inherit}{U_2} & \style{font-family:inherit}{U_3} \\\hline \style{font-family:inherit}{P(X\cap Y)} & 0 & \frac14 & \frac14 \\[0pt]\hline \style{font-family:inherit}{P(X)P(Y)} & \frac14\times\frac12=\frac18 & \frac14\times\frac34=\frac38 & \frac12\times\frac12=\frac14 \\[0pt]\hline \style{font-family:inherit}{ X\text{ and }Y\text{ are}\ldots} & \textbf{dependent} & \textbf{dependent} & \textbf{independent} \\[0pt]\hline \style{font-family:inherit}{X\text{ and }Y\text{ are}\ldots} & \textbf{mutually exclusive} & \textbf{not ME} & \textbf{not ME} \end{array}\hskip-5.5pt \end{array} $$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$enter image description here


The mistake I did in probability was when I was first introduced to un-equally likely outcomes. The example was something like this-

A toin is tossed one time. If it appears head, a die is rolled. The sample space is $S = (H1 , H2 , H3 , H4 , H5 , H6 , T)$

After reading upon the sample space, I blindly thought that all of these outcomes are equally likely and probability of any ( say $H1$) is $1/7$. However I realised the mistake later.


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