Four fair dice are thrown independently 27 times. Then the expected number of times, at least two di

Q: Wait, why do we use Binomial Distribution here? Aren't the dice independent?

That is exactly why we use it! The Binomial Distribution is the perfect mathematical model for a sequence of independent trials where each trial has only two outcomes: success (getting a 3 or 5) or failure (getting anything else). Since each die is independent, the probability of success remains constant at $p = 1/3$ for every die.

Q: Can I just calculate $P(X=0)$ and $P(X=1)$ and subtract from 1?

Absolutely! In fact, that is often the 'pro' move. Calculating $P(X \ge 2)$ directly requires summing three terms, whereas the complement method only requires two. Both paths lead to the same result, $11/27$. It is a great way to double-check your work during the exam.

Q: Why is the expected value just $N \times P$?

This is a fundamental property of the Binomial Distribution. If you repeat an experiment with probability $p$ of success $N$ times, the expected number of successes is defined as $E[X] = Np$. It is the weighted average of all possible outcomes, and it simplifies beautifully here because $N=27$ and $p=11/27$.

Q: Is it possible to get a 3 or 5 on more than 4 dice?

No, because we are only throwing 4 dice at a time. The variable $X$ (the number of successes) is bounded by the number of trials $n=4$. That is why our sum stops at $P(X=4)$.

Q: What if I made a mistake in calculating the probability of a single die?

Always re-read the question carefully! It says 'a three or a five'. There are 6 faces on a die, so the favorable outcomes are 2. Thus, $p = 2/6 = 1/3$. If you accidentally used $1/6$, the entire chain of calculations would collapse. Always verify your base probability before diving into the binomial formula.

Question

Four fair dice are thrown independently 27 times. Then the expected number of times, at least two dice show up a three or a five, is

Sigma Percentile · Accepted Answer

### The Dance of Probability: Unlocking the Mystery of the Dice

Imagine you are sitting in a quiet room, a set of four fair dice in your hand. You are about to embark on a journey of 27 trials. In each trial, you roll all four dice at once.

You are looking for a specific pattern: at least two of those dice must show a 3 or a 5. It sounds like a game of chance, but in the world of JEE Advanced, chance is just another word for a beautiful, predictable mathematical structure.

Let us break this down, step by step, and see how the chaos of the dice yields to the elegance of the Binomial Distribution.

### Phase 1: The Single Die Perspective

Before we look at the group of four, we must understand the individual. A fair die has six faces: $\{1, 2, 3, 4, 5, 6\}$.

We are interested in the event where a die shows a 3 or a 5. These are our 'success' outcomes. Since there are two favorable outcomes out of six total, the probability of success for a single die is $p = \frac{2}{6} = \frac{1}{3}$.

Consequently, the probability of failure—getting anything else—is $q = 1 - p = \frac{2}{3}$. This simple ratio is the heartbeat of our entire problem.

### Phase 2: The Binomial Framework

Now, we throw four dice. Because each die is independent, we are looking at a Binomial Distribution with $n = 4$ trials.

We define $X$ as the number of dice showing a 3 or a 5. We want the probability that $X \ge 2$. This means we are interested in the scenarios where we get exactly 2, 3, or 4 successes.

Mathematically, we express this as:

$$P(X \ge 2) = P(X=2) + P(X=3) + P(X=4)$$

Using the binomial formula $P(X=r) = \binom{n}{r} p^r q^{n-r}$, we can calculate each piece of the puzzle.

### Phase 3: Calculating the Probabilities

First, for $X=2$:

$$P(X=2) = \binom{4}{2} \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^2 = 6 \cdot \frac{1}{9} \cdot \frac{4}{9} = \frac{24}{81}$$

Next, for $X=3$:

$$P(X=3) = \binom{4}{3} \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^1 = 4 \cdot \frac{1}{27} \cdot \frac{2}{3} = \frac{8}{81}$$

Finally, for $X=4$:

$$P(X=4) = \binom{4}{4} \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^0 = 1 \cdot \frac{1}{81} \cdot 1 = \frac{1}{81}$$

When we sum these up, we get the total probability for a single trial of four dice:

$$P(X \ge 2) = \frac{24+8+1}{81} = \frac{33}{81}$$

Simplifying this fraction by dividing both numerator and denominator by 3, we arrive at $P(X \ge 2) = \frac{11}{27}$.

### Phase 4: The Grand Expectation

We have performed the hard work. We know that in any single roll of four dice, the probability of our desired event is $\frac{11}{27}$.

The problem asks for the expected number of times this happens over 27 trials. The expectation $E$ of a binomial process is simply the number of trials $N$ multiplied by the probability of success $p$:

$$E = N \cdot P(X \ge 2) = 27 \cdot \frac{11}{27} = 11$$

And there it is! The complexity vanishes, leaving behind the clean, integer result of **11**.

It is a reminder that even when the dice seem to roll randomly, the underlying laws of probability are always working in perfect harmony. Keep this clarity with you as you tackle your next challenge—you have the tools, the logic, and the persistence to master any problem.

Animated Solution for Mathematics - Probability: Four fair dice are thrown independently 27 times. Then the expected number of times, at least two dice show up a three or a five, is

Enter Numerical Value:

Visualized Solution

Single Die Probability

Defining a Single Trial

Binomial Distribution Setup

Calculating $P (X = 2)$

Calculating $P (X = 3)$

Calculating $P (X = 4)$

Total Probability $P (X \geq 2)$

Final Expected Value $E$

The Sigma Insight: Random Variables and Probability Distributions

The Dance of Probability

Phase 1

Phase 2

Phase 3

Phase 4

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