ISS 2016 Statistics Paper-1 Solution: Question 24 (Conditional Probability and Independence)
Continuing our question-by-question walkthrough of the ISS Statistics Paper-1 archive, this post picks up right where the last one left off. We are working strictly in order, one previous-year question at a time, so that every concept gets the full step-by-step treatment instead of a rushed one-line answer.
Today's question comes from ISS 2016, Statistics Paper-1, Question 24, and it tests one of the most foundational — and most frequently misunderstood — ideas in probability theory: the relationship between conditional probability, statistical independence, and mutual exclusivity.
Quick Summary
Topic:Conditional probability, independent events, and mutually exclusive events
Question Reference:ISS 2016, Statistics Paper-1, Question 24
Correct Answer:(a) 1, 2 and 3 only
The Question
A and B are two events such that P(A) = P(A | B) = 1/4 and P(B | A) = 1/2. Which of the following are correct?
A and B are independent events.
P(A) < P(B)
P(A̅ | B) = 3/4
A and B are mutually exclusive events.
Select the correct answer using the code given below:
(a) 1, 2 and 3 only
(b) 1, 2 and 4 only
(c) 2, 3 and 4 only
(d) 1, 3 and 4 only
Step 1: Write Down What Is Given
We are told three numbers:
P(A) = 1/4
P(A | B) = 1/4
P(B | A) = 1/2
Everything in this question follows from these three facts, so before touching any statement, it helps to convert the conditional probabilities into a statement about the intersection P(A ∩ B), since that is the quantity every formula in probability ultimately depends on.
Step 2: Test Statement 1 — Are A and B Independent?
Recall the definition of independence: two events A and B (with P(B) > 0) are independent if and only if P(A | B) = P(A). This is not a coincidence or a shortcut — it is literally the definition, because conditioning on B should not change the probability of A if the two events don't influence each other.
Here we are directly given:
P(A | B) = 1/4 = P(A)
Since P(A | B) equals P(A) exactly, A and B satisfy the definition of independence. So Statement 1 is TRUE.
Step 3: Find P(B) Using the Second Piece of Information
Now use P(B | A) = 1/2. By the definition of conditional probability:
P(B | A) = P(A ∩ B) / P(A)
So P(A ∩ B) = P(B | A) × P(A) = (1/2) × (1/4) = 1/8.
Now go back to the conditional probability formula for P(A | B):
P(A | B) = P(A ∩ B) / P(B)
Substituting the values we know:
1/4 = (1/8) / P(B)
Rearranging, P(B) = (1/8) ÷ (1/4) = (1/8) × 4 = 1/2.
So we now have P(A) = 1/4 and P(B) = 1/2.
Pro Tip: Whenever a question gives you two conditional probabilities like P(A | B) and P(B | A), the fastest and safest route is to convert both into expressions for P(A ∩ B) using the basic definition P(X | Y) = P(X ∩ Y) / P(Y), then equate them. This turns the whole problem into simple algebra instead of trying to "reason" about independence intuitively, which is exactly where most students slip up under exam pressure. Drilling this conversion step until it becomes automatic — writing out P(A∩B) first, every single time, before touching any statement — is a habit that is emphasized heavily in structured problem-solving practice, and it is one of the fastest ways to stop losing marks on "which of the following is/are correct" questions.
Step 4: Test Statement 2 — Compare P(A) and P(B)
We now know P(A) = 1/4 = 0.25 and P(B) = 1/2 = 0.5. Clearly, 1/4 < 1/2, so P(A) < P(B) holds.Statement 2 is TRUE.
Step 5: Test Statement 3 — The Complement Rule
Statement 3 claims P(A̅ | B) = 3/4, where A̅ denotes "not A" (the complement of A).
There is a very useful and completely general rule here: for any event B with P(B) > 0,
P(A | B) + P(A̅ | B) = 1
This is because, once we condition on B, the sample space effectively becomes B itself, and within that reduced sample space, "A happens" and "A does not happen" are complementary events that must add up to 1 — this is true whether or not A and B are independent.
We already know P(A | B) = 1/4, so:
P(A̅ | B) = 1 − P(A | B) = 1 − 1/4 = 3/4
This matches Statement 3 exactly, so Statement 3 is TRUE.
Step 6: Test Statement 4 — Are A and B Mutually Exclusive?
Two events are mutually exclusive if they cannot occur together, which means P(A ∩ B) = 0. But in Step 3 we calculated:
P(A ∩ B) = 1/8 ≠ 0
Since the intersection has a positive, non-zero probability, A and B clearly can occur together, so they are not mutually exclusive.Statement 4 is FALSE.
It's worth pausing on why this makes intuitive sense too: independent events with strictly positive probabilities (P(A) > 0 and P(B) > 0 here) can never be mutually exclusive, except in the trivial edge case. If A and B were mutually exclusive, then knowing B occurred would tell you A definitely did not occur, which would make P(A | B) = 0 — but we were given P(A | B) = 1/4, not 0. So Statement 4 being false is consistent with everything else we found.
Final Answer
Statements 1, 2 and 3 are true; Statement 4 is false. This corresponds exactly to option (a) 1, 2 and 3 only, which matches the official ISS 2016 answer key for Question 24.
Why This Question Matters
This question is a compact test of three ideas that appear constantly throughout the ISS syllabus: the formal definition of conditional probability, the precise definition of independence (not just an intuitive notion of it), and the distinction between independence and mutual exclusivity — two concepts students very commonly confuse. In reality, independence and mutual exclusivity sit at almost opposite ends of a spectrum: mutually exclusive events with positive probability are about as far from independent as two events can get, because knowing one occurred tells you everything about the other (it didn't happen). Questions like this one are a reliable way for examiners to check whether a candidate has internalized the algebra behind these definitions rather than just memorizing the words.
Frequently Asked Questions
What is the difference between independent events and mutually exclusive events?
Independent events are ones where the occurrence of one does not affect the probability of the other, formally P(A | B) = P(A). Mutually exclusive events are ones that cannot both happen at the same time, formally P(A ∩ B) = 0. If both events have positive probability, they can never be both independent and mutually exclusive at the same time — these are two genuinely different, almost opposing, properties.
Why does P(A | B) = P(A) prove independence?
This is literally the definition of independence. Intuitively, if knowing that B occurred doesn't change how likely A is to occur, then A's occurrence does not depend on B in any probabilistic sense. Once you confirm this equality holds, you can also derive P(A ∩ B) = P(A) × P(B), which is the more commonly quoted (but equivalent) test for independence.
How do I quickly find P(A ∩ B) from conditional probabilities?
Use the basic definition directly: P(A ∩ B) = P(A | B) × P(B), or equivalently P(A ∩ B) = P(B | A) × P(A). Whichever side of the conditional you are given, multiply the conditional probability by the probability of the event you conditioned on.
Does P(A | B) + P(A̅ | B) = 1 always hold, even if A and B are not independent?
Yes, always, as long as P(B) > 0. This is because conditioning on B restricts you to a smaller sample space (the outcomes where B occurs), and within that restricted space, A and "not A" still partition all outcomes completely, so their conditional probabilities must add to 1 regardless of any relationship between A and B.
Can two independent events also be mutually exclusive?
Only in the trivial case where at least one of them has probability 0. If both P(A) > 0 and P(B) > 0, independent events must have P(A ∩ B) = P(A)P(B) > 0, which directly contradicts the mutual exclusivity condition P(A ∩ B) = 0. So for any events with genuinely positive probability, independence and mutual exclusivity are mutually incompatible properties.
What is the most common mistake students make on this type of question?
The most common mistake is assuming mutual exclusivity and independence are somehow related or that one implies the other. Many students, upon seeing "A and B are independent" is true, incorrectly assume mutual exclusivity might also hold, or vice versa. Always check mutual exclusivity separately by directly computing P(A ∩ B) and comparing it to zero, rather than relying on intuition.
If any step above felt unclear, or if you solved this one differently, drop a comment below with your doubt — and if you found this walkthrough useful, do share it with a fellow ISS aspirant who might be working through the same paper.
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