ISS PREVIOUS YEAR 2016 PAPER-1 SET-A Q.no.- 13

ISS PREVIOUS YEAR 2016 PAPER-1 SET-A

ISS Previous year question 2016

Question 13

Exponential Distribution Problem (Bulb Lifetime)

The lifetime of a bulb follows an exponential distribution with mean 100 hours. The bulb is switched on for exactly 4 hours every day and remains switched off for the remaining time.

What is the probability that the bulb stops working on or before the 25th day?

Options:

(a) (1 − e⁻¹) / (1 − e⁻¹⁄²⁵) (b) 1 − e⁻¹⁄²⁵ (c) 1 − e⁻¹ (d) e⁻¹

Solution

This is a classic exponential distribution + real-life interpretation problem, very important for ISS and other exams.

Step 1: Understand the Distribution

Let X be the lifetime of the bulb measured in working hours.

We are given:

Mean lifetime = 100 hours

For exponential distribution:

Mean = 1 / λ

So,

λ = 1 / 100

Step 2: Total Working Time in 25 Days

The bulb is ON for 4 hours per day.

So, in 25 days:

Total working time = 25 × 4 = 100 hours

Step 3: Required Probability

The bulb fails on or before the 25th day means:

It fails within 100 working hours

So we need:

P(X ≤ 100)

Step 4: Use Exponential Distribution Formula

For exponential distribution:

P(X ≤ x) = 1 − e^(−λx)

Substitute values:

P(X ≤ 100) = 1 − e^(−(1/100 × 100))

Step 5: Simplify

P(X ≤ 100) = 1 − e⁻¹

Final Answer

1 − e⁻¹

Correct option: (c)

Concept Insight (Very Important)

• Lifetime is measured in working time, not calendar time

• Even though 25 days pass, the bulb actually works only 100 hours

• Exponential distribution depends only on total working time

Quick Exam Trick

Whenever exponential lifetime is given with ON/OFF condition:

Convert everything into effective working time first

Then apply:

P(X ≤ x) = 1 − e^(−λx)

Final Thought

“Smart students don’t solve more…They interpret better.”

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