Three dice are rolled. If the probability of getting different numbers on the three dice is p/q, whe

Q: Wait, why do we treat the dice as distinct? Aren't they identical?

That is a classic point of confusion! In probability, we treat the dice as distinct (e.g., Die 1, Die 2, Die 3) to ensure that every outcome in our sample space is equally likely. If we treated them as identical, the outcome $(1, 1, 2)$ would be less likely than $(1, 2, 3)$, making the standard probability formula $P(A) = \frac{n(A)}{n(S)}$ invalid. By treating them as distinct, we ensure each of the $216$ outcomes has a probability of $\frac{1}{216}$.

Q: Can I use combinations instead of the multiplication principle?

Absolutely! You can choose $3$ distinct numbers out of $6$ in $^6C_3$ ways, and then arrange those $3$ numbers in $3!$ ways. So, $n(A) = ^6C_3 \times 3! = 20 \times 6 = 120$. Both methods are mathematically equivalent and perfectly valid. The multiplication principle is often faster, but combinations provide a great way to verify your work.

Q: What if the question asked for 'at least two dice showing the same number'?

Great question! Whenever you see 'at least,' think about the complement. The complement of 'at least two dice showing the same number' is 'all three dice showing different numbers.' Since we already calculated the probability of all different numbers as $\frac{5}{9}$, the probability of the complement is simply $1 - \frac{5}{9} = \frac{4}{9}$.

Question

Three dice are rolled. If the probability of getting different numbers on the three dice is $\frac{p}{q}$, where $p$ and $q$ are co-prime, then $q - p$ is equal to

Options:

(A) 2

(B) 1

(C) 3

(D) 4

Sigma Percentile · Accepted Answer

### The Beauty of Probability: Rolling Three Dice

Imagine you are holding three standard six-sided dice. You are about to roll them, and you want to know the likelihood that every single one lands on a different number.

This is not just a game; it is a beautiful exercise in counting and logic. Let us break this down step by step.

### The Foundation: The Sample Space

Before we can talk about winning, we must understand the entire universe of possibilities. We call this the sample space, denoted as $n(S)$.

When you roll one die, there are $6$ possible outcomes. When you roll three, because each die is independent, we use the multiplication principle.

$$n(S) = 6 	imes 6 	imes 6 = 6^3 = 216$$

This is the total number of ways the dice can land.

### The Constraint: The Logic of Distinctness

Now, we look for the favorable outcomes, $n(A)$, where all three dice show different numbers. Let the outcomes be $d_1, d_2, d_3$. We require $d_1 
eq d_2 
eq d_3$.

Think of this as filling three empty slots. For the first die, $d_1$, there are no restrictions; it can be any of the $6$ faces. So, we have $6$ choices.

For the second die, $d_2$, we have a constraint: it must be different from $d_1$. Since one number is already taken, we have $6 - 1 = 5$ choices left.

Finally, for the third die, $d_3$, it must be different from both $d_1$ and $d_2$. Since two distinct numbers are already occupied, we have $6 - 2 = 4$ choices remaining.

### The Calculation: Bringing it Together

To find the total number of favorable outcomes, we multiply these choices:

$$n(A) = 6 	imes 5 	imes 4 = 120$$

We have $120$ ways to get three distinct numbers. Now, the probability $P(A)$ is simply the ratio of favorable outcomes to total outcomes:

$$P(A) = \frac{n(A)}{n(S)} = \frac{120}{216}$$

### The Final Polish: Simplification

We are almost there. We need to simplify the fraction $\frac{120}{216}$.

Both numbers are divisible by $24$. Dividing the numerator by $24$ gives $5$, and dividing the denominator by $24$ gives $9$.

$$P(A) = \frac{5}{9}$$

The problem states this is $\frac{p}{q}$ where $p$ and $q$ are co-prime. Since $5$ and $9$ share no common factors, we have $p=5$ and $q=9$.

The final step is to calculate **$q - p = 9 - 5 = 4$**.

You have just navigated the logic of independent events and constraints with elegance. Keep this mindset, and no probability problem will ever intimidate you again.

Animated Solution for Mathematics - Probability: Three dice are rolled. If the probability of getting different numbers on the three dice is $\frac{p}{q}$ , where $p$ and $q$ are co-prime, then $q - p$ is equal to

Select Answer:

Visualized Solution

Visualizing the Experiment

Total Possible Outcomes $n (S)$

Calculating $n (S)$

Condition for Favorable Outcomes

Choices for the First Die

Choices for the Second Die

Choices for the Third Die

Calculating Favorable Outcomes $n (A)$

Probability Formula

Simplifying the Probability

Comparing with $\frac{p}{q}$

Final Calculation: $q - p$

The Sigma Insight: Classical Definition of Probability

The Beauty of Probability

The Foundation

The Constraint

The Calculation

The Final Polish

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