Probability Interview Questions

Sample Answer

The answer is $\binom{16}{6} - \binom{10}{6} - 6\binom{10}{5}$. Count all teams from 16 people, then subtract teams with 0 engineers, which is choosing all 6 from the 10 researchers, and subtract teams with exactly 1 engineer, which is choose 1 of 6 engineers and 5 of 10 researchers. This avoids case overlap because the excluded cases, 0 engineers and exactly 1 engineer, are disjoint.

Sample Answer

You could do direct placement of buys into gaps, or use inclusion-exclusion on adjacent buy pairs. Direct placement wins here because the separation constraint translates to choosing slots. First arrange the 7 sells in $7!$ ways, which creates 8 gaps around them, then choose 5 of those 8 gaps to place the buys, and permute buys within the chosen gaps. The count is $$7!\,\binom{8}{5}\,5!.$$

Sample Answer

You could do it with Bayes directly, or with a quick probability tree. Bayes is fastest here because it is one line: $P(B\mid +)=\frac{P(+\mid B)P(B)}{P(+\mid A)P(A)+P(+\mid B)P(B)}$. Compute $\frac{0.6\cdot 0.3}{0.1\cdot 0.7+0.6\cdot 0.3}=\frac{0.18}{0.25}=0.72$. The common mistake is swapping $P(+\mid B)$ with $P(B\mid +)$.

Sample Answer

Let me define events: $B$ is the user was in variant B, and $C$ is the user converted. You want $P(B\mid C)$, so Bayes says $P(B\mid C)=\frac{P(C\mid B)P(B)}{P(C\mid A)P(A)+P(C\mid B)P(B)}$. Here $P(B)=P(A)=0.5$, so it is $\frac{0.05\cdot 0.5}{0.04\cdot 0.5+0.05\cdot 0.5}=\frac{0.05}{0.09}=\frac{5}{9}\approx 0.556$. Intuitively, converters are biased toward the higher converting arm, but not overwhelmingly because the lift is small.

Sample Answer

This question is checking whether you can define events cleanly, use complements, and condition in the right direction. Let $N$ be network drop and $D$ be decoder bug, independent, and alert $A$ is $N\cup D$. You want $P(D\mid A)=\frac{P(D)}{P(N\cup D)}$ because $D\subseteq A$, and $P(N\cup D)=P(N)+P(D)-P(N)P(D)=0.01+0.005-0.00005=0.01495$. So $P(D\mid A)=\frac{0.005}{0.01495}\approx 0.334$.

Sample Answer

Rate 120 per minute means on average 2 updates per second. In a Poisson process, the interarrival time $T$ is exponential with parameter $\lambda=2$ per second, so $T \sim \text{Exponential}(2)$. You want $P(T>1)$, and for an exponential that is $e^{-\lambda t}$. Plugging in gives $P(T>1)=e^{-2\cdot 1}=e^{-2}$.

Sample Answer

This question is checking whether you can map a large number of independent Bernoulli trials to the right count model and then choose a workable approximation. Exactly, $X \sim \text{Binomial}(n=50000,p=0.02)$ with mean $\mu=np=1000$ and variance $\sigma^2=np(1-p)=980$. For a tail like $P(X\ge 1100)$, you would use a normal approximation with continuity correction: $$P(X\ge 1100)\approx P\left(Z\ge \frac{1099.5-1000}{\sqrt{980}}\right).$$ If $p$ were smaller you might also consider a Poisson($\lambda=np$) approximation, but here the normal is typically better because $np$ and $n(1-p)$ are both large.

Sample Answer

This question is checking whether you can avoid summing a negative binomial pmf and instead use additivity on i.i.d. geometric waiting times. Write $T=\sum_{k=1}^r G_k$ where each $G_k$ is the number of flips to get the next head, i.i.d. geometric with mean $1/p$ and variance $(1-p)/p^2$. Then $$\mathbb{E}[T]=r\,\mathbb{E}[G_1]=\frac{r}{p},\quad \mathrm{Var}(T)=r\,\mathrm{Var}(G_1)=\frac{r(1-p)}{p^2}.$$ Independence across head-to-head blocks is the key justification.

Sample Answer

The standard move is Hoeffding since you are bounded, it gives a clean exponential tail without knowing the variance. But here, $X_i\in[0,1]$ matters because the bound is dimension-free: $$\Pr\left(|\bar X-\mu|\ge \varepsilon\right)\le 2\exp(-2n\varepsilon^2),$$ so it suffices to take $$n\ge \frac{1}{2\varepsilon^2}\log\frac{2}{\delta}.$$ You switch to a CLT approximation when you need a sharper constant and you have enough data to estimate $\sigma^2=\mathrm{Var}(X_i)$ reliably, because Hoeffding can be conservative when $\sigma^2\ll 1$.

Sample Answer

Get this wrong in production and you ship an experiment decision with a false sense of precision, so you need a bound that is valid without asymptotics. The right call is Chebyshev on $\hat p=\frac{1}{n}\sum X_i$ with $\mathrm{Var}(\hat p)=p(1-p)/n\le p/n$. Chebyshev gives $$\Pr(|\hat p-p|\ge 0.1p)\le \frac{\mathrm{Var}(\hat p)}{(0.1p)^2}\le \frac{p/n}{0.01p^2}=\frac{100}{np}.$$ Set this to $\le 0.01$ to get $n\ge 10000/p$, conservative but interview-correct under the stated tool constraints.

Sample Answer

The standard move is to write a recursion for the expected hitting time via one step lookahead. But here, the absorbing boundary conditions matter because they pin the recursion with exact values at 0 and 2. Let $E_i$ be the expected time to absorption from $i$, then $E_0=E_2=0$ and $$E_1=1+\tfrac12E_0+\tfrac12E_2=1.$$ So you absorb in expected time 1 because you must leave state 1 in the first step and you always land in an absorbing state.

Sample Answer

Get this wrong in production and you can end up with an agent that keeps exploring forever, blowing up regret and making offline evaluation meaningless. The right call is to ensure the target set you want to hit is reached with probability 1 from all starting states, and in a finite chain that is implied if the target set is reachable from every state and there are no other closed communicating classes outside it. Then you compute expected hitting times by solving the linear system $h(i)=1+\sum_j P_{ij}h(j)$ for $i$ outside the target with boundary $h=0$ on the target, or you upper bound using a uniform lower bound on progress, for example a geometric bound if you can show a per step hit probability at least $p>0$. In a finite irreducible chain, expected return times are finite and $\mathbb{E}[\tau]<\infty$ for hitting any nonempty set.

Sample Answer

$h(i)=\mathbb{P}_i(\text{hit 4 before 0})$ sounds reasonable to treat as linear in $i$, but that breaks under drift when $p\neq 1/2$. Plugging in a linear guess can accidentally satisfy the recursion only in the symmetric case. A brute force simulation does not work because the interviewer wants the exact setup and boundary handling. That leaves solving the boundary value problem: for $i=1,2,3$, $$h(i)=p\,h(i+1)+(1-p)\,h(i-1),$$ with $h(0)=0$, and the reflecting condition gives $h(4)=h(3)$ since from 4 you immediately go to 3. Solve the resulting 3 equation system for $h(2)$.

Sample Answer

Most candidates default to applying optional stopping blindly, but that fails here because you must check a condition like $\mathbb{E}[\tau]<\infty$ or uniform integrability of $M_{t\wedge\tau}$. For the simple symmetric walk and two sided bounded barriers, $\tau$ has finite expectation, so you can apply optional stopping to $S_t$ itself, which is a martingale. You get $\mathbb{E}[S_\tau]=\mathbb{E}[S_0]=0$, so $a\,\mathbb{P}(S_\tau=a) + (-b)\,\mathbb{P}(S_\tau=-b)=0$. With probabilities summing to 1, you solve $$a p - b(1-p)=0\Rightarrow p=\frac{b}{a+b}.$$