Conditional Probability Under Sample Space Reduction
You have a set $B$ with $|B| = n$ elements, and a subset $A \subseteq B$ with $|A| = k$. You draw a uniformly random element from $B$.
For any element $b \in B \setminus A$, what is the probability that $b$ was selected, given that the selected element is not in $A$? Derive the general formula, then evaluate it for $n = 10$, $k = 4$.
Hints
- Conditioning on $x \notin A$ is equivalent to restricting attention to the $n - k$ elements outside $A$. How does that change the sample space?
- Apply the definition of conditional probability: $P(x = b \mid x \notin A) = P(x = b,\, x \notin A) / P(x \notin A)$.
- Since $b \notin A$, the joint event $\{x = b\} \cap \{x \notin A\}$ simplifies to just $\{x = b\}$, giving numerator $1/n$. The denominator is $1 - k/n$.
Worked Solution
How to Think About It: The heuristic is *conditioning a uniform distribution on a subset gives the uniform distribution on that subset*. Being told "the draw is not in $A$" throws away $k$ of the $n$ equally likely outcomes and cannot break the symmetry among the $n - k$ survivors — no element of $B \setminus A$ is favored over any other, so each must get probability $1/(n-k)$. There is no real trap here, but the interviewer wants to see you know *why* the shortcut is legal: the mechanical reason is renormalization — every surviving outcome keeps its original mass $1/n$ and you divide by the total surviving mass $(n-k)/n$.
Quick Estimate: For $n = 10$, $k = 4$: the element $b$ starts with mass $\tfrac{1}{10}$; the conditioning event "not in $A$" contains $10 - 4 = 6$ elements, so it has mass $\tfrac{6}{10}$; renormalizing, $b$ gets $\tfrac{1/10}{6/10} = \tfrac{1}{6} \approx 0.167$. Symmetry cross-check: 6 equally-likely survivors must each get $\tfrac16$ — same number, two independent one-line routes.
Approach: Definition of conditional probability, using that $\{x = b\}$ is a sub-event of $\{x \notin A\}$.
Formal Solution:
Let $x$ denote the selected element. We want $P(x = b \mid x \notin A)$ for a fixed $b \in B \setminus A$.
By the definition of conditional probability: $$P(x = b \mid x \notin A) = \frac{P(x = b,\, x \notin A)}{P(x \notin A)}$$
Numerator. Since $b \notin A$, the event $\{x = b\}$ already implies $\{x \notin A\}$, so the joint event is just $\{x = b\}$: $$P(x = b,\, x \notin A) = P(x = b) = \frac{1}{n}$$
Denominator. $A$ contains $k$ of the $n$ equally likely elements: $$P(x \notin A) = 1 - \frac{k}{n} = \frac{n - k}{n}$$
Combine: $$P(x = b \mid x \notin A) = \frac{1/n}{(n-k)/n} = \boxed{\frac{1}{n-k}}$$
Since this value is the same for every $b \in B \setminus A$ and there are $n - k$ such elements summing to $1$, the conditional distribution is exactly uniform on $B \setminus A$ — confirming the symmetry intuition.
Answer: $P(x = b \mid x \notin A) = \dfrac{1}{n-k}$ in general; for $n = 10$, $k = 4$ this is $\boxed{\dfrac{1}{6}}$.
Intuition
Conditioning on an event is just restricting your universe. When you learn the outcome is not in $A$, you effectively throw out $k$ possibilities and renormalize. The remaining $n - k$ elements are still equally likely relative to each other -- uniform distributions stay uniform after conditioning on a symmetric event like 'not in $A$.' So the conditional probability is just $1/(n-k)$.
This pattern shows up constantly in quant work. In market microstructure, conditioning on a trade being informative (or not) shrinks the effective set of counterparty types you are dealing with. In filtering and Bayesian updating, conditioning on an observation is always a sample-space restriction followed by renormalization. Getting comfortable with this mechanical step -- joint probability over marginal probability -- is the foundation for everything from Bayes' theorem to the change-of-measure techniques used in derivatives pricing.