The probability, that in a randomly selected 3-digit number at least two digits are odd, is

Q: Why can't I just calculate the number of ways to pick 2 odd digits and multiply by 3? It feels like overcounting.

That is a common trap! If you just pick 2 odd digits and 1 even digit and multiply by 3, you are assuming all positions are identical. But the first position is restricted (it cannot be zero). By splitting into $OOE$, $OEO$, and $EOO$, you handle the zero restriction explicitly for each case, ensuring you don't accidentally include invalid numbers like $023$.

Q: Why is the $EOO$ case different from $OOE$?

Because of the '3-digit number' definition! In $OOE$, the first digit is odd, so it can be $1, 3, 5, 7,$ or $9$. None of these are zero. But in $EOO$, the first digit is even. If you blindly pick from $\{0, 2, 4, 6, 8\}$, you might pick $0$, which makes the number a 2-digit number (e.g., $013$ is just $13$). We must exclude zero for the first position.

Q: Is there a faster way using the complement?

Yes, you could calculate the total minus the cases with 0 or 1 odd digits. However, for JEE Advanced, direct counting is often safer because it allows you to verify your logic step-by-step. The complement method is elegant but prone to 'off-by-one' errors if you aren't careful with the zero restriction. Stick to the method that gives you the most confidence!

Question

The probability, that in a randomly selected 3-digit number at least two digits are odd, is

Options:

(A) $\frac{19}{36}$

(B) $\frac{15}{36}$

(C) $\frac{13}{36}$

(D) $\frac{23}{36}$

Sigma Percentile · Accepted Answer

### The Architecture of Probability

Welcome, future engineer. Today, we are not just solving a probability problem; we are building a mental model of how to navigate constraints in combinatorics.

When you look at a 3-digit number, do not just see a value. See a structure consisting of three slots: the Hundreds, the Tens, and the Units. This is your playground.

### Phase 1: The Sample Space

Before we hunt for our favorable outcomes, we must define the universe. We are looking at 3-digit numbers, which range from $100$ to $999$.

The first slot, the Hundreds place, is the gatekeeper. It cannot be $0$, or else we fall into the realm of 2-digit numbers. Thus, it has $9$ choices ($\{1, 2, \dots, 9\}$).

The Tens and Units places are free spirits; they can be any digit from $0$ to $9$, giving us $10$ choices each. By the Fundamental Principle of Counting, our total sample space $n(S)$ is:

$$n(S) = 9 	imes 10 	imes 10 = 900$$

This is our denominator.

### Phase 2: The Logic of 'At Least'

'At least two odd digits' is a phrase that demands precision. It means we are satisfied if we have exactly two odd digits, or if we have all three.

We must partition this into mutually exclusive cases to ensure we count every valid number exactly once. We have two sets of digits: Odd $\{1, 3, 5, 7, 9\}$ (count of $5$) and Even $\{0, 2, 4, 6, 8\}$ (count of $5$).

### Phase 3: The Cases

Let us begin with the simplest scenario: all three digits are odd. This is the $OOO$ pattern.

For each slot, we have $5$ choices. Thus, the number of ways is:

$$5 	imes 5 	imes 5 = 125$$

Now, we move to the 'exactly two odd' scenarios. We have three patterns: $OOE$, $OEO$, and $EOO$.

For $OOE$ (Odd-Odd-Even), the first two slots are odd ($5 	imes 5$) and the last is even ($5$). That gives us $125$ ways. For $OEO$ (Odd-Even-Odd), the logic is identical: $5 	imes 5 	imes 5 = 125$ ways.

### Phase 4: The Trap

Now, we arrive at the $EOO$ (Even-Odd-Odd) pattern. This is where the JEE examiner tests your vigilance.

If we place the even digit at the front, we must remember the gatekeeper rule: the first digit cannot be $0$. Therefore, our even choices for the first slot are restricted to $\{2, 4, 6, 8\}$, giving us only $4$ choices.

The remaining two slots are odd, giving us $5 	imes 5$ choices. Thus, the number of ways is:

$$4 	imes 5 	imes 5 = 100$$

If you had used $5$ here, you would have included numbers like $013$, which are not 3-digit numbers!

### The Final Synthesis

We have navigated the traps. Now, we sum our favorable outcomes $n(E)$:

$$n(E) = 125 + 125 + 125 + 100 = 475$$

The probability $P(E)$ is:

$$P(E) = \frac{475}{900}$$

By dividing both the numerator and the denominator by $25$, we arrive at the elegant result:

$$\mathbf{P(E) = \frac{19}{36}}$$

You have successfully mastered the constraints and conquered the counting. Keep this vigilance in your toolkit for every exam you face.

Animated Solution for Mathematics - Probability: The probability, that in a randomly selected 3-digit number at least two digits are odd, is

Select Answer:

Visualized Solution

Visualizing the Sample Space

Defining the Event: At Least Two Odd

Case 1: All Three Digits are Odd

Case 2a: Exactly Two Odd (Pattern OOE)

Case 2b: Exactly Two Odd (Pattern OEO)

Case 2c: The Trap (Pattern EOO)

Summing Favorable Outcomes

Final Probability Calculation

The Sigma Insight: Classical Definition of Probability

The Architecture of Probability

Phase 1

Phase 2

Phase 3

Phase 4

The Final Synthesis

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