ISS 2016 Statistics Paper-1 Solution: Question 13 (Exponential Bulb Lifetime Usage)

Continuing our question-by-question walk through the ISS 2016 Statistics Paper-1 paper, we now arrive at Question 13 — a neat probability puzzle that mixes the exponential distribution with a bit of everyday reasoning about how a bulb is actually used. If you've been following this series from Question 1, you'll notice the pattern by now: read the question slowly, identify exactly what quantity is random, and let the algebra follow.

This one trips up a lot of aspirants not because the mathematics is hard, but because of a subtle modelling choice buried in the wording. Let's unpack it fully.

Quick Summary

• Topic:Exponential distribution — CDF, memoryless property, and modelling "usage time" versus calendar time

• Question Reference:ISS 2016, Statistics Paper-1, Question 13

• Correct Answer:Option (c),1 − e⁻¹

The Question, As Asked

The lifetime of a bulb is exponentially distributed with mean 100 hours. The bulb remains switched on for exactly 4 hours every day and remains switched off the remaining time. What is the probability that the bulb stops working on or before the 25th day?

• (a) (1 − e⁻¹) / (1 − e−1/25)

• (b) 1 − e−1/25

• (c) 1 − e⁻¹

• (d) e⁻¹

Step 1: Identify What Is Actually Random

Start by writing down what we are told. Let T denote the lifetime of the bulb, and we're told T follows an exponential distribution with mean 100 hours. In symbols:

T ~ Exponential(λ), with E(T) = 1/λ = 100, so λ = 1/100.

Now here is the part that decides the whole question: what does "lifetime" actually measure — calendar time since the bulb was first installed, or the total number of hours the bulb has actually been lit? Physically, a bulb's filament degrades only while current is flowing through it — it doesn't wear out while sitting switched off. So the 100-hour mean lifetime is measured in hours of actual usage (on-time), not in calendar hours or days.

Step 2: Convert "25th Day" Into Total On-Time

The bulb is switched on for exactly 4 hours every day. So by the end of the 25th day, the total number of hours the bulb has actually been running is:

Total on-time by day 25 = 25 days × 4 hours/day = 100 hours.

So the event "the bulb stops working on or before the 25th day" is exactly the same event as "the bulb's total usage lifetime T is less than or equal to 100 hours." We have translated a statement about calendar days into a statement about the random variable T itself — which is the only way to actually use the exponential distribution we were given.

Pro Tip: The trickiest part of this question is realizing that the on/off pattern (4 hours on, 20 hours off, repeated) is a complete red herring once you've converted days into total on-hours. Because the bulb only "consumes" its lifetime while lit, and because we only care about the cumulative on-time, it makes no difference whether those 100 hours of usage happened in one continuous stretch or were spread across 25 separate 4-hour sessions. The way this kind of question is usually taught is to immediately strip away the daily on/off schedule and ask: "how many actual hours of use have accumulated?" — once you have that number, the problem collapses into a single, ordinary exponential CDF calculation.

Step 3: Apply the Exponential CDF

For an exponential random variable T with rate parameter λ, the cumulative distribution function (the probability that T is less than or equal to some value t) is:

F(t) = P(T ≤ t) = 1 − e−λt, for t ≥ 0.

We want P(T ≤ 100), with λ = 1/100. Substitute:

F(100) = 1 − e−(1/100) × 100 = 1 − e−1.

Notice how cleanly the numbers cancel: the mean lifetime (100 hours) and the total accumulated on-time by day 25 (also 100 hours) are exactly equal, so λt = 1, and we're simply left with 1 − e⁻¹.

Step 4: Match to the Options and Confirm

Our derived answer, 1 − e⁻¹, is exactly option (c). Checking the official ISS 2016 Statistics Paper-1 answer key, Question 13 is indeed marked as option C — so our independent derivation matches the official key perfectly. There's no discrepancy to flag here.

It's worth noting why the other options are tempting-but-wrong distractors: option (b), 1 − e−1/25, is what you'd get if you mistakenly used t = 25 (calendar days) directly as if it were measured in the same units as the mean (100 hours), without converting to actual on-time hours. Option (a) looks like a conditional-probability expression, which isn't needed here since there's no additional conditioning event. Option (d), e⁻¹, is simply the complement of the correct answer — the probability the bulb is still working after 100 hours of use, not the probability it has failed.

Why This Question Matters

This question is a good test of whether you understand what the exponential distribution's parameter actually represents, rather than just memorizing its CDF formula. In real applications — reliability engineering, insurance claim modelling, queueing theory — the "clock" that a lifetime distribution runs on is very often not the same as the calendar clock, and recognizing which clock applies is half the battle. Aspirants who drill this kind of unit-conversion reasoning until it becomes automatic tend to move noticeably faster through the reliability and life-testing questions that show up almost every year in the ISS syllabus, since the underlying trick — convert everything into the units the distribution's parameter is defined on — repeats itself again and again with different dressing.

Frequently Asked Questions

Why do we use 100 total on-hours instead of 25 × 24 = 600 calendar hours?

Because the exponential distribution here models the bulb's usage lifetime — how many hours of actual illumination it can sustain before failing — not how many hours of calendar time pass around it. A bulb switched off does not age, so only the 4 on-hours per day count toward its 100-hour mean lifetime budget.

What is the memoryless property, and why does it matter for this question?

The memoryless property of the exponential distribution says that the distribution of remaining lifetime doesn't depend on how much time has already elapsed, provided the bulb hasn't failed yet. In this question, it justifies why we can simply add up the 4-hour on-periods across 25 separate days and treat the total as equivalent to one continuous 100-hour stretch of usage — the gaps in between (when the bulb is off) don't reset or otherwise affect the failure clock.

How do we know λ = 1/100 from "mean 100 hours"?

For an exponential distribution with rate parameter λ, the mean is always 1/λ. So if we're told the mean is 100 hours, we solve 1/λ = 100, giving λ = 1/100 per hour. This λ is what goes into the CDF formula 1 − e−λt.

What's the general formula students should memorize here?

For T ~ Exponential(λ), P(T ≤ t) = 1 − e−λt and P(T > t) = e−λt. It helps to also remember that λ = 1/(mean), so you can always rewrite the exponent as t/(mean) directly, without separately solving for λ first.

Would the approach change if the daily usage were, say, 5 hours instead of 4?

Yes — you'd simply recompute the total accumulated on-time as 25 × 5 = 125 hours, and then evaluate F(125) = 1 − e−125/100 = 1 − e−1.25 instead. The method — convert days into total usage hours, then plug into the exponential CDF — stays exactly the same.

What's the single most common mistake students make on this type of question?

Plugging the number of days (25) directly into the CDF as if it were measured in the same units as the mean (which is in hours), without first converting to actual usage hours. That mistake produces option (b) here, which is exactly why it's included as a distractor.

If anything here felt unclear, or you approached the problem differently, drop a comment below — working through the discrepancy is often more instructive than the original solution. And if you know a fellow ISS aspirant working through the same paper, feel free to share this with them; going through these one question at a time really does add up.