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: Recall the memoryless property of the exponential distribution.
The exponential distribution has a special property called memorylessness: for any \(s, t \geq 0\), \(P(X > s+t \mid X > s) = P(X > t)\). This says that once you know \(X\) has already exceeded \(s\), the chance it goes on to exceed \(s+t\) more is the same as the chance a fresh copy of \(X\) exceeds \(t\) from zero. It does not matter how long you have already waited.

Step 2: Match the numbers to the property.
We want \(P(X > 10 \mid X > 5)\). Here \(s = 5\) and \(s + t = 10\), so \(t = 5\). By memorylessness,
\[ P(X > 10 \mid X > 5) = P(X > 5+5 \mid X > 5) = P(X > 5) \]

Step 3: Substitute the given value.
We are told \(P(X > 5) = 0.35\), so the conditional probability equals 0.35 directly, with no need to even find \(\lambda\).

Step 4: Verify using the exponential formula.
For \(X\) exponential, \(P(X > x) = e^{-x/\lambda}\). From \(P(X>5) = e^{-5/\lambda} = 0.35\), squaring gives \(P(X > 10) = e^{-10/\lambda} = (e^{-5/\lambda})^2 = 0.35^2 = 0.1225\). Since \(X > 10\) automatically means \(X > 5\),
\[ P(X > 10 \mid X > 5) = \frac{P(X > 10)}{P(X > 5)} = \frac{0.1225}{0.35} = 0.35 \]
This matches Step 3, confirming the answer.

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