ISS 2016 Statistics Paper-1 Solution: Question 12 (Exponential Lifetime Comparison)
Continuing our question-by-question walkthrough of the ISS 2016 Statistics Paper-1, this post takes up Question 12 — a nice practical problem on comparing two independent exponential lifetimes. If you've been following along since Question 1, you'll notice this one leans on integration rather than pure algebra, so we'll go slowly through every step.
Quick Summary
Topic: Exponential distribution – probability that one of two independent lifetimes is smaller
Question Reference: ISS 2016 Statistics Paper-1, Question 12
Correct Answer: Option (b) – 5/9
The Question
To compare the lifetimes of bulbs produced by two companies, A and B, one bulb of each company was selected at random and their lifetimes were observed. Assume that the bulbs have exponentially distributed lifetimes with mean 1000 days and 800 days for company A and company B, respectively. What is the probability that the bulb from company B fails first?
(a) 4/9
(b) 5/9
(c) 25/81
(d) 25/41
Step 1: Set Up the Two Random Variables
Let X denote the lifetime (in days) of the bulb from company A, and Y denote the lifetime of the bulb from company B. Both are given to be exponentially distributed, and they are independent of each other since the two bulbs were selected separately.
For an exponential distribution with mean μ, the rate parameter λ is simply 1/μ. This is a definition worth memorizing cold: mean and rate are reciprocals of each other for the exponential distribution.
So here:
X ~ Exponential with mean 1000, so λA = 1/1000
Y ~ Exponential with mean 800, so λB = 1/800
The probability density functions are:
fX(x) = λA e−λAx, for x > 0
fY(y) = λB e−λBy, for y > 0
Step 2: Translate the Question into a Probability Statement
"The bulb from company B fails first" means Y is smaller than X, i.e. we need P(Y < X).
Since X and Y are independent continuous random variables, we can write this using the survival function of X. The survival function P(X > y) tells us the probability that bulb A is still working at time y. For an exponential variable, P(X > y) = e−λAy — this itself comes directly from integrating the density of X from y to infinity, so it's worth remembering as a standalone fact rather than re-deriving it every time.
To find P(Y < X), we condition on the value that Y takes. If Y = y, then we need X to be larger than y, which happens with probability e−λAy. We then average this over all possible values of y, weighted by the density of Y:
P(Y < X) = ∫0∞ fY(y) · P(X > y) dy = ∫0∞ λB e−λBy · e−λAy dy
Step 3: Combine the Exponents and Integrate
Since e−λBy · e−λAy = e−(λA+λB)y, the integral becomes:
P(Y < X) = λB ∫0∞ e−(λA+λB)y dy
Now recall the basic integration fact: ∫0∞ e−ky dy = 1/k for any constant k > 0 (this is just the total-probability integral of an exponential density with rate k, divided by k, or equivalently you can do it directly: the antiderivative of e−ky is −(1/k)e−ky, and evaluating from 0 to ∞ gives 0 − (−1/k) = 1/k).
Applying this with k = λA + λB:
P(Y < X) = λB · [1 / (λA + λB)] = λB / (λA + λB)
This is a clean, general result worth remembering: for two independent exponential variables with rates λA and λB, the probability that the "B" variable is the smaller one equals λB divided by the sum of the two rates.
Pro Tip: The trickiest part of this question isn't the calculus – it's converting "mean 1000 days" and "mean 800 days" into rates without mixing them up. A very common slip is plugging the means directly into the ratio formula instead of the rates (getting 1000/1800 instead of the correct rate-based ratio). The way this is usually taught is to immediately write down λ = 1/mean for every exponential variable the moment you read the question, before touching any other algebra — that one habit, drilled until it's automatic, eliminates almost every careless error on this type of problem.
Step 4: Substitute the Actual Rates
We have λA = 1/1000 and λB = 1/800. Let's find a common denominator to add them. The LCM of 1000 and 800 is 4000.
λA = 1/1000 = 4/4000
λB = 1/800 = 5/4000
So:
λA + λB = 4/4000 + 5/4000 = 9/4000
Now substitute into our formula:
P(Y < X) = λB / (λA + λB) = (5/4000) / (9/4000)
The 4000 in the numerator and denominator cancels out directly, leaving:
P(Y < X) = 5/9
Step 5: Final Answer and Confirmation
The probability that the bulb from company B fails first is 5/9, which is option (b). This matches the official ISS 2016 answer key exactly (Question 12 is listed as B in the key), so we can be confident in both the setup and the arithmetic.
As a sanity check, notice that company B's bulbs have a shorter mean lifetime (800 days versus 1000 days), so it makes intuitive sense that B is more likely to fail first – and 5/9 is indeed greater than 1/2, confirming the direction of our answer is correct even before checking the exact fraction.
Why This Question Matters
This question tests a genuinely useful and frequently recurring idea in reliability theory and survival analysis: given two independent exponential "waiting times," what is the chance one beats the other? This exact formula – λB/(λA+λB) – shows up again when you study the minimum of several independent exponential variables (a key building block in reliability engineering, queuing theory, and competing risks models), so getting comfortable with this derivation now pays off well beyond this one MCQ. Students preparing with Sunrise Classes often see this same integral reappear in disguised forms in later papers, which is exactly why working through the full derivation – rather than memorizing just the final ratio – is time well spent.
Frequently Asked Questions
Why is the rate parameter equal to 1 divided by the mean for an exponential distribution?
For an exponential distribution with density f(x) = λe−λx, the mean E(X) works out to 1/λ when you compute the integral ∫0∞ xλe−λxdx (using integration by parts). So if you're told the mean is 1000, you simply invert it to get λ = 1/1000. This relationship is fixed for every exponential variable, so it is always safe to use directly.
Could this problem be solved without integration?
Yes. Because the minimum of two independent exponential variables is itself exponential with rate λA+λB, and each of the two original variables is equally likely (in a weighted sense) to be the one that achieves that minimum, the probability that Y is the smaller one is proportional to its own rate: λB/(λA+λB). This is a standard result you can quote directly once you've derived it once, though showing the derivation is safer in an exam that asks for justification.
What does "memoryless" mean for the exponential distribution, and is it used here?
The memoryless property says that P(X > s+t | X > s) = P(X > t) – in plain words, a bulb that has already survived s days is no more or less likely to fail in the next t days than a brand-new bulb. It isn't directly invoked in this particular derivation, but it is the same underlying property that makes the survival function P(X > y) = e−λy so simple to work with, which is exactly what we used in Step 2.
What if the question had asked for the probability that bulb A fails first instead?
You would use the exact same formula with the roles swapped: P(X < Y) = λA/(λA+λB). In this problem that would give 4/9. Notice that 5/9 + 4/9 = 1, which is a useful self-check – the two probabilities must add up to 1 since exactly one bulb fails first (ties have zero probability for continuous variables).
Is there a common mistake students make with this type of question?
The most frequent error is using the means (1000 and 800) directly in the ratio instead of converting to rates first, which changes both the numerator and denominator in a way that quietly flips the answer. Always convert every mean to a rate as your very first step before writing any probability expression.
Where else in the ISS syllabus does this concept reappear?
This same idea of comparing or combining independent exponential variables underlies reliability system questions (series and parallel systems of components), the distribution of order statistics from exponential samples, and certain Poisson process inter-arrival time problems, all of which show up regularly across ISS Paper-1 papers.
If anything in this derivation felt unclear, drop a comment below and we'll walk through it further – and if you found this useful, do share it with a fellow ISS aspirant who might be revising the same topic.
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