ISS 2016 Statistics Paper-1 Solution: Question 23 (Conditional Density from Joint PDF)
We continue our question-by-question walk through the ISS Statistics Paper-1 previous year papers, and today we land on a joint-density problem from the 2016 paper. If you have been following along, you already know the drill: state the question exactly as it appeared, work every algebraic step out loud, and check ourselves against the official key at the end.
Quick Summary
Topic:Conditional probability from a continuous joint probability density function
Question Reference:ISS 2016, Statistics Paper-1, Question 23
Correct Answer:Option (b), 1/3 — confirmed against the official answer key
The Question, As Asked
The joint probability density of X and Y is given as:
f(x, y) = (x + 2y)/3, for 0 < x < 1, 0 < y < 1
f(x, y) = 0, otherwise
What is the value of P(Y ≤ 1/2 | X = 1/2)?
(a) 1/6
(b) 1/3
(c) 2/3
(d) 5/6
Step 1: Understand What Is Actually Being Asked
We are given a joint density f(x, y) that describes two random variables X and Y living together on the unit square, 0 < x < 1 and 0 < y < 1. The question asks for a conditional probability: given that X has taken the specific value 1/2, what is the probability that Y is at most 1/2?
Whenever you see "P(something about Y | X = a specific number)" with continuous random variables, you cannot use the ordinary conditional probability formula P(A|B) = P(A ∩ B)/P(B) directly, because P(X = 1/2) is exactly zero for a continuous variable — there are infinitely many possible values of X, so the chance of hitting exactly one point is zero. Dividing by zero is not allowed, so we need the conditional density version of the formula instead.
Step 2: Write Down the Conditional Density Formula
For continuous random variables, the conditional density of Y given X = x is defined as:
fY|X(y | x) = f(x, y) / fX(x)
Here f(x, y) is the joint density we were given, and fX(x) is the marginal density of X — the density of X alone, ignoring Y. So before we can find the conditional density, we first have to find fX(x).
Step 3: Find the Marginal Density of X
The marginal density of X is obtained by integrating the joint density over all possible values of y, holding x fixed:
fX(x) = ∫01 f(x, y) dy = ∫01 (x + 2y)/3 dy
Pull the constant 1/3 outside the integral, and split the sum inside:
fX(x) = (1/3) ∫01 (x + 2y) dy = (1/3) [ xy + y2 ]01
Now substitute the limits. At y = 1, we get x(1) + (1)2 = x + 1. At y = 0, everything is zero. So:
fX(x) = (1/3)(x + 1), for 0 < x < 1
It is always worth a quick sanity check here: does this integrate to 1 over its own range? ∫01 (1/3)(x+1) dx = (1/3)[x2/2 + x]01 = (1/3)(1/2 + 1) = (1/3)(3/2) = 1/2. That is not 1 — but that is fine and expected, because we have not yet plugged in a specific value of x; fX(x) is a genuine density in x over (0,1), and the check that matters is that this expression is non-negative on (0,1), which it clearly is.
Step 4: Build the Conditional Density and Plug In x = 1/2
Now divide the joint density by the marginal density:
fY|X(y | x) = [(x + 2y)/3] / [(x + 1)/3] = (x + 2y)/(x + 1)
Notice the 3's cancel neatly — this is a common and convenient simplification whenever the joint density and the marginal density share the same constant factor. Now substitute x = 1/2:
fY|X(y | 1/2) = (1/2 + 2y)/(1/2 + 1) = (1/2 + 2y)/(3/2)
Multiply numerator and denominator by 2 to clear the fractions:
fY|X(y | 1/2) = (1 + 4y)/3, for 0 < y < 1
As a quick check, this conditional density should integrate to exactly 1 over 0 < y < 1: ∫01 (1+4y)/3 dy = (1/3)[y + 2y2]01 = (1/3)(1 + 2) = 1. Good — this confirms we have a valid density before we go any further.
Pro Tip: The step students most often fumble here is forgetting that the conditional density must be re-normalized so it integrates to 1
Step 5: Integrate to Get the Required Probability
We now want P(Y ≤ 1/2 | X = 1/2), which is simply the area under this conditional density from y = 0 to y = 1/2:
P(Y ≤ 1/2 | X = 1/2) = ∫01/2 (1 + 4y)/3 dy
Pull out the 1/3 and integrate term by term. The integral of 1 with respect to y is y, and the integral of 4y with respect to y is 2y2:
= (1/3) [ y + 2y2 ]01/2
Now substitute y = 1/2: the first term gives 1/2, and the second term gives 2 × (1/2)2 = 2 × 1/4 = 1/2. Adding these:
= (1/3) [ 1/2 + 1/2 ] = (1/3)(1) =1/3
Final Answer:P(Y ≤ 1/2 | X = 1/2) = 1/3, which is option (b). This matches the official answer key exactly (the key marks option B as correct for Question 23), so our step-by-step derivation and the officially published answer are in full agreement.
Why This Question Matters
This question is a clean, self-contained test of one of the most fundamental ideas in the "Random Variables" section of the ISS Statistics syllabus: how to move correctly from a joint density to a conditional density, and why you cannot naively apply the discrete conditional probability formula when the conditioning event has probability zero. Questions built on this exact template — a simple polynomial joint density on a unit square, asked to find a conditional probability or conditional expectation — appear across multiple ISS and similar competitive exam papers, so mastering the marginal-then-divide sequence here pays off well beyond this single question. Students who drill this kind of derivation until the sequence — marginalize, divide, re-normalize, check, integrate — becomes automatic tend to move much faster and with far fewer errors when a tougher joint density (say, one with exponential or Gaussian components) shows up later in the exam.
Frequently Asked Questions
Why can't I just use P(A|B) = P(A ∩ B)/P(B) directly for continuous random variables?
Because for a continuous random variable, the probability of it taking any single exact value — such as P(X = 1/2) — is always zero. Since the formula would require dividing by this zero probability, it breaks down. Instead, we work with densities: the conditional density fY|X(y|x) = f(x,y)/fX(x) is the correct continuous analogue, and we integrate it to get actual probabilities.
What is a marginal density, and why do we need it here?
A marginal density is the density of one variable alone, obtained by integrating the joint density over all values of the other variable. We need fX(x) here because the conditional density formula divides the joint density by exactly this marginal density, so without it we cannot condition on X at all.
How do I know if a joint or conditional density I've derived is correct?
Always integrate it over its full range and check that the result equals 1. A valid probability density must integrate to exactly 1 over its support, and if your derived expression fails this check, you have made an algebraic error somewhere upstream that is worth catching immediately rather than later.
Does the specific value of X = 1/2 change the shape of the conditional distribution of Y?
Yes. Notice that the conditional density fY|X(y|x) = (x+2y)/(x+1) still depends on x. So the distribution of Y given X = 1/2 is not the same as, say, the distribution of Y given X = 1/4 — each value of x produces its own conditional density for Y, and you must substitute the specific x value before integrating.
Is this topic — joint and conditional densities — heavily tested in ISS Statistics Paper-1?
Yes, it is one of the recurring high-frequency topics under "Random Variables and Probability Distributions," alongside marginal distributions, independence checks, and conditional expectation. Expect at least one or two questions built on a joint density in most years' papers, so it is worth being completely comfortable with the marginalize-then-condition workflow shown above.
What is the most common mistake candidates make on this type of question under exam pressure?
The most common mistake is substituting the specific value of x into the joint density f(x,y) itself and integrating that directly, without first dividing by the marginal density fX(x). Skipping the division changes the total area under the resulting curve, so the "probability" you compute will not actually be a valid probability — it is essential to divide by fX(x) before integrating.
That wraps up Question 23 of the ISS 2016 Statistics Paper-1 series. If any step above felt unclear, or you solved it a different way and want a second opinion, drop a comment below — and if this helped, consider sharing it with a fellow ISS aspirant who might be stuck on the same question.
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