Question

Let \(X\) be an exponentially distributed random variable with mean \(\lambda (> 0)\). If \(P(X > 5) = 0.35\), then the conditional probability \(P(X > 10 \mid X > 5)\) is ________. (Rounded off to two decimal places)

Hint:

Use the memoryless property of the exponential distribution: P(X>10|X>5) reduces to P(X>5).

Answer 1

Correct Answer: 0.35

Step 1: Write the survival function of the exponential distribution.
If $X$ is exponential with mean $\lambda$, its survival function is $P(X > x) = e^{-x/\lambda}$ for $x \geq 0$. This directly gives the chance $X$ exceeds any value $x$.

Step 2: Write the conditional probability as a ratio.
Since the event $X > 10$ is a subset of the event $X > 5$ (if $X$ is bigger than 10 it is automatically bigger than 5),
\[ P(X > 10 \mid X > 5) = \frac{P(X > 10 \text{ and } X > 5)}{P(X > 5)} = \frac{P(X > 10)}{P(X > 5)} \]

Step 3: Express both survival probabilities using the exponential formula.
$P(X > 5) = e^{-5/\lambda}$ and $P(X > 10) = e^{-10/\lambda} = \left(e^{-5/\lambda}\right)^2$. So the ratio becomes
\[ \frac{P(X>10)}{P(X>5)} = \frac{\left(e^{-5/\lambda}\right)^2}{e^{-5/\lambda}} = e^{-5/\lambda} = P(X > 5) \]
This shows the ratio always collapses back to $P(X > 5)$ for this exponential setup, which is really just the memoryless property appearing algebraically.

Step 4: Plug in the given value.
We are given $P(X > 5) = 0.35$, so
\[ P(X > 10 \mid X > 5) = 0.35 \]

Final Answer:
\[ \boxed{0.35} \]