ISS 2016 Statistics Paper-1 Solution: Question 20 (Conditional Probability with Uniform Distribution)

Continuing our question-by-question walk through the ISS Statistics Paper-1 previous year papers, today we pick up right where we left off with Question 20 of the 2016 paper. If you've been following this series from Question 1, this one is a nice change of pace — no distributions with tricky parameters, just plain conditional probability built on the humble uniform distribution.

Quick Summary

• Topic: Conditional probability using the Uniform(0, 1) distribution

• Question Reference: ISS 2016, Statistics Paper-1, Question 20

• Correct Answer: Option (a),(b − a) / (2 − a − b)

The Question

If X has a uniform distribution over the interval (0, 1), then for any real numbers a and b, 0 < a < b < 1, what is

P[a < X < b | X > (a + b)/2]

equal to?

• (a) (b − a) / (2 − a − b)

• (b) 2(b − a) / (b + a)

• (c) 2(b − a) / (2 − a − b)

• (d) (b − a) / (b + a)

Step 1: Write Down What the Uniform(0, 1) Distribution Actually Gives You

Before touching the conditional probability, it helps to be completely clear on what "X is uniform on (0, 1)" means in practice. It means X can take any value between 0 and 1, and every sub-interval of the same length is equally likely. Formally, the density function is

f(x) = 1, for 0 < x < 1, and f(x) = 0 otherwise.

The one fact you will use again and again for a Uniform(0, 1) variable is this: for any interval (c, d) that lies inside (0, 1), the probability that X falls in that interval is simply the length of the interval:

P(c < X < d) = d − c.

That single fact is really all the "distribution theory" this question needs — the rest is careful bookkeeping with the conditional probability formula.

Step 2: Recall the Definition of Conditional Probability

For any two events A and B (with P(B) > 0), the conditional probability of A given B is defined as

P(A | B) = P(A ∩ B) / P(B).

Here, take:

• A = the event {a < X < b}

• B = the event {X > (a + b)/2}

So we need two things: the probability of the intersection A ∩ B, and the probability of B on its own. Both are just lengths of intervals, once we figure out exactly which interval each event corresponds to.

Step 3: Name the Midpoint and Locate It

To keep the algebra clean, let m = (a + b)/2. This is just the midpoint of a and b. Since we are told 0 < a < b < 1, the midpoint of any two distinct numbers always lies strictly between them, so

a < m < b.

This one observation is the entire key to the question. Once you know where m sits relative to a and b, the intersection in Step 4 becomes almost automatic instead of something you have to puzzle over.

Pro Tip: Whenever a conditional probability question gives you a "compound" condition like X > (a + b)/2 sitting inside a range like a < X < b, the very first thing to do — before any probability calculation — is to work out where that boundary point actually falls on the number line relative to a and b. This is exactly the habit that's drilled into students early on: draw a quick number line with a, m, and b marked on it, and the intersection of the two events becomes something you can just read off, rather than something you have to reason about abstractly.

Step 4: Work Out the Intersection A ∩ B

A ∩ B is the set of values of X that satisfy both a < X < b AND X > m at the same time. Since we've just shown a < m < b, the numbers between a and m already fail the second condition (they are not greater than m), so they drop out. What survives is exactly the stretch from m up to b:

A ∩ B = {m < X < b}.

This is the step where a lot of students go wrong — it's tempting to assume the intersection is just the whole interval (a, b) again, since "a < X < b" was already given. But the extra condition X > m genuinely shrinks the interval on the left-hand side, from a up to m. Missing this is exactly how you'd end up at option (c) instead of the correct option (a), as we'll see explicitly at the end.

Step 5: Compute the Two Probabilities

Using the "probability equals length" fact from Step 1:

Numerator — P(A ∩ B) = P(m < X < b):

P(m < X < b) = b − m = b − (a + b)/2.

To simplify, write b as 2b/2 so both terms share a denominator:

b − (a + b)/2 = (2b)/2 − (a + b)/2 = (2b − a − b)/2 = (b − a)/2.

So P(A ∩ B) = (b − a)/2.

Denominator — P(B) = P(X > m):

P(X > m) = 1 − m = 1 − (a + b)/2.

Again, write 1 as 2/2:

1 − (a + b)/2 = 2/2 − (a + b)/2 = (2 − a − b)/2.

So P(B) = (2 − a − b)/2.

Step 6: Divide to Get the Final Answer

Now apply the conditional probability formula from Step 2:

P(A | B) = P(A ∩ B) / P(B) = [(b − a)/2] / [(2 − a − b)/2].

The 2's in the numerator and denominator are identical, so they cancel completely, leaving:

P(A | B) = (b − a) / (2 − a − b).

This is exactly option (a). If instead you had (incorrectly, as flagged in Step 4) kept the full interval (a, b) as your intersection rather than shrinking it to (m, b), your numerator would have come out as (b − a) instead of (b − a)/2, and dividing by the same denominator would land you on 2(b − a)/(2 − a − b) — which is precisely the distractor option (c). Seeing exactly how a distractor is manufactured like this is one of the more useful things you can take away from an MCQ paper, since it tells you exactly which shortcut to be suspicious of next time.

Cross-checking with the official ISS 2016 Statistics Paper-1 (Series A) answer key: Question 20 is listed as A, which matches our derivation exactly.

Why This Question Matters

On the surface this looks like a "just plug into a formula" question, but it's really testing whether you can carefully translate a conditional probability statement into the right sub-interval before doing any arithmetic. This exact skill — locating where a conditioning boundary falls relative to the event you care about — shows up constantly in ISS and other statistics exams, whether the underlying distribution is uniform, exponential, or something more exotic. Get comfortable with the "draw the number line first" habit here, on the simplest possible distribution, and it transfers directly to much harder conditional probability and reliability-type questions later in the same paper.

Frequently Asked Questions

Why is P(c < X < d) simply equal to d − c for a Uniform(0, 1) variable?

Because the density function of a Uniform(0, 1) variable is constant and equal to 1 over the interval (0, 1). Integrating a constant height of 1 over an interval of width (d − c) gives exactly (d − c), so probability and interval length become the same number for this particular distribution.

Why does the intersection of {a < X < b} and {X > m} become {m < X < b} and not something else?

An intersection keeps only the values that satisfy both conditions simultaneously. Since a < m < b, any value of X between a and m fails the condition X > m, so it is excluded. Only the values between m and b satisfy both conditions at once, which is why the intersection shrinks to (m, b).

Is the midpoint (a + b)/2 always guaranteed to lie between a and b?

Yes. For any two distinct real numbers a and b with a < b, their average (a + b)/2 always lies strictly between them. This is a basic algebraic fact and is exactly what makes Step 3 of this solution valid without any extra checking.

What if the question had asked for P(X < a | X < m) instead?

You would follow the exact same method: first locate where a sits relative to m (here a < m, since a is one of the two numbers averaged to get m), find the intersection — which would just be {X < a} since that is the smaller, fully contained interval — and then divide its length by the length of {X < m}, i.e., a / m.

How can I quickly avoid falling for a distractor like option (c) in this type of question?

Always write out the intersection event explicitly as an interval before computing any probability, rather than trying to combine the two events "in your head." Once the intersection is written down as (m, b) on paper, it becomes very hard to accidentally use the wrong interval length in the numerator.

Does this method change if X is uniform on some other interval, like (0, 10) instead of (0, 1)?

The method is identical — you would still express both probabilities as interval lengths and divide. The only difference is that the "probability equals length" shortcut would instead be "probability equals length divided by the total length of the interval," i.e., you would divide each length by 10 instead of by 1. Since the same scaling factor appears in both the numerator and denominator here, it would actually cancel out either way.

That wraps up Question 20 of the ISS 2016 Statistics Paper-1. If any step above felt unclear, or you solved it differently, drop a comment below — and if you know a fellow ISS aspirant working through the same paper, feel free to share this with them so we can keep going through the rest of the paper together.