Q.A person observed the first 4 digits of your 6-digit PIN. What is the probability that the person can guess your PIN?
(A) 1 81
(B) 1 100
(C) 1 90
(D) 1 1 ASSERTION-REASON BASED QUESTIONS (Question numbers 19 and 20 are Assertion -Reason based questions carrying 1 mark each. Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the options (A), (B), (C) and (D) as given below.) (A) Both (A) and (R) are true and (R) is the correct explanation of (A). (B) Both (A) and (R) are true but (R) is not the correct explanation of (A). (C) (A) is true but (R) is false. (D) (A) is false but (R) is true.

Concept: Conditional Probability – the probability of guessing the remaining digits correctly given that the first 4 are already known.

Step 1: A 6-digit PIN has $10^6$ possible combinations (each digit 0–9).

Step 2: The person already knows the first 4 digits, so only the last 2 digits are unknown.

Step 3: The last 2 digits can be any of $10 \times 10 = 100$ possibilities. Only 1 of these is correct. …

The problem is about conditional probability — but in a very natural, everyday sense. The person has observed the first 4 digits of your 6-digit PIN. That means those 4 digits are no longer uncertain; they are known. The only uncertainty lies in the remaining 2 digits.

Why does this matter? Because probability is always about what you don’t know. Once information is given, the space of possibilities shrinks. Here, the observation reduces the problem from guessing a full 6-digit PIN (which would have $10^6 = 1,000,000$ possibilities) to guessing just the last 2 digits.

Let’s walk through it step by step.

1. Understand what is known and what is unknown.

   A 6-digit PIN has each digit chosen from 0 to 9 — that’s 10 choices per digit. The person already knows the first 4 digits. So those are fixed. The unknown part is only the last 2 digits.

2. Count the number of possible outcomes for the unknown part.

   The last 2 digits are independent of each other. The first unknown digit can be any of 10 digits, and the second unknown digit can also be any of 10 digits. So the total number of equally likely possibilities for the last 2 digits is:

3. Count the number of favourable outcomes. There is exactly one correct PIN — the one that matches your actual last 2 digits. So only 1 outcome out of the 100 is the right one. …

Method: Probability of a Correct Guess When Part of the Answer Is Already Known

Use this whenever some information is already revealed and only a remaining part is left to guess.

Steps

Step 1: Separate the known part from the unknown part.

Revealed information is no longer uncertain, so it drops out of the count. Only the still-unknown part carries any probability. This is conditional thinking: the given information shrinks the sample space.

Step 2: Count the equally-likely possibilities for the unknown part. …

Common Mistakes

Mistake 1: Guessing all six digits instead of only the unknown two.

Why it's wrong: the first four digits were observed, so they are known, not guessed — treating the whole PIN as unknown gives $\frac{1}{10^6}$. Correct approach: only the last two digits are uncertain, so the count is $10\times10=100$ and the probability is $\frac{1}{100}$. …