About this question
Easy · Probability & Statistics · Quant Trader interview question · probability, geometric-distribution, memoryless, conditional-probability
Let $X$ be a geometric random variable representing the number of trials until the first success, with success probability $p$ on each trial. The geometric distribution is memoryless. This means that knowing you've already had $m$ failures doesn't change the probability of needing $n$ more trials. Express $P(X > m+n | X > m)$ in terms of $n$. Remember that $P(A|B) = \frac{P(A \cap B)}{P(B)}$.