The mean and variance of a binomial distribution are α and α/3 respectively. If P(X = 1) = 4/243, th

Q: Wait, why do we divide Variance by Mean? Is that a standard trick or just a coincidence?

It is a classic, elegant trick! In a Binomial Distribution, the Mean is $np$ and the Variance is $npq$. By dividing Variance by Mean, the $n$ and $p$ terms cancel out perfectly, leaving you with just $q$. It is the fastest way to extract the probability of failure from the given parameters.

Q: Can I use logarithms to solve for $n$ in the equation $\frac{2n}{3^n} = \frac{4}{3^5}$?

You absolutely could, but in the JEE Advanced exam, time is your most precious resource. Whenever you see powers of the same base (like $3^n$ and $3^5$), try to compare the structures first. Testing integer values for $n$ is often much faster and less prone to calculation errors than using logs.

Q: Why do we add $P(X=4)$ and $P(X=5)$? Why not multiply them?

Great question! We add them because the events '$X=4$' and '$X=5$' are mutually exclusive—the random variable $X$ cannot be both 4 and 5 at the same time. In probability, when we want the union of mutually exclusive events, we sum their individual probabilities.

Question

The mean and variance of a binomial distribution are $\alpha$ and $\frac{\alpha}{3}$ respectively. If $P(X = 1) = \frac{4}{243}$, then $P(X = 4 	ext{ or } 5)$ is equal to:

Options:

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

(B) $\frac{64}{81}$

(C) $\frac{16}{27}$

(D) $\frac{145}{243}$

Sigma Percentile · Accepted Answer

### The Architecture of Chance: Decoding the Binomial Distribution

Welcome, future engineer. Today, we are not just solving a probability problem; we are peeling back the layers of a Binomial Distribution to reveal the hidden parameters that define it. In the world of JEE Advanced, problems like this are not tests of your ability to memorize formulas, but tests of your ability to see the 'DNA' of a mathematical structure.

### Phase 1: The DNA of the Distribution

Imagine you are handed a black box labeled 'Binomial Distribution'. You are told two things: the Mean is $\alpha$ and the Variance is $\frac{\alpha}{3}$.

Recall the fundamental definitions for a Binomial Distribution $B(n, p)$:

1. The Mean is defined as $E[X] = np = \alpha$.

2. The Variance is defined as $Var(X) = npq = \frac{\alpha}{3}$.

Here, $n$ is the number of trials, $p$ is the probability of success, and $q = 1 - p$ is the probability of failure. If we divide the Variance by the Mean, we isolate the probability of failure:

$$\frac{npq}{np} = \frac{\frac{\alpha}{3}}{\alpha}$$

The $n$ and $p$ terms cancel out entirely, leaving us with $q = \frac{1}{3}$. Consequently, the probability of success must be $p = 1 - q = \frac{2}{3}$.

### Phase 2: The Detective Work

Now that we know $p = \frac{2}{3}$ and $q = \frac{1}{3}$, we must determine $n$, the number of trials. The problem provides the clue $P(X = 1) = \frac{4}{243}$.

We invoke the general formula for binomial probability:

$$P(X = r) = \binom{n}{r} p^r q^{n-r}$$

Substituting $r = 1$, $p = \frac{2}{3}$, and $q = \frac{1}{3}$, we obtain:

$$P(X = 1) = \binom{n}{1} \left(\frac{2}{3}ight)^1 \left(\frac{1}{3}ight)^{n-1} = \frac{2n}{3^n}$$

Equating this to the given value $\frac{4}{243}$, we observe that $243 = 3^5$:

$$\frac{2n}{3^n} = \frac{4}{3^5}$$

By inspection, we see that $n = 6$ satisfies the equation, as $\frac{2(6)}{3^6} = \frac{12}{729} = \frac{4}{243}$. Thus, we have determined that **$n = 6$**.

### Phase 3: The Final Calculation

We now calculate $P(X = 4 	ext{ or } 5)$. Since these are mutually exclusive events, we sum their individual probabilities:

$$P(X = 4 	ext{ or } 5) = P(X = 4) + P(X = 5)$$

First, we calculate $P(X = 4)$:

$$P(X = 4) = \binom{6}{4} \left(\frac{2}{3}ight)^4 \left(\frac{1}{3}ight)^2 = 15 \cdot \frac{16}{81} \cdot \frac{1}{9} = \frac{240}{729}$$

Next, we calculate $P(X = 5)$:

$$P(X = 5) = \binom{6}{5} \left(\frac{2}{3}ight)^5 \left(\frac{1}{3}ight)^1 = 6 \cdot \frac{32}{243} \cdot \frac{1}{3} = \frac{192}{729}$$

Summing these values yields:

$$P(X = 4 	ext{ or } 5) = \frac{240 + 192}{729} = \frac{432}{729}$$

Simplifying the fraction by dividing both the numerator and denominator by 27, we arrive at the final result:

**$$P(X = 4 	ext{ or } 5) = \frac{16}{27}$$**

### Conclusion

We started with abstract variables and, through logical deduction, uncovered the specific parameters of a system. This is the essence of mathematics in the JEE Advanced curriculum—it is not about the final number, but the clarity of the path taken to reach it. You have mastered the binomial distribution today.

Animated Solution for Mathematics - Probability: The mean and variance of a binomial distribution are $α$ and $\frac{α}{3}$ respectively. If $P (X = 1) = \frac{4}{243}$ , then $P (X = 4 or 5)$ is equal to:

Select Answer:

Visualized Solution

Given Parameters

Mean and Variance Formulas

Relating Variance to Mean

Calculating the Probability $p$

The $P (X = 1)$ Formula

Substituting Known Values

Setting up the Equation for $n$

Solving for the Number of Trials $n$

Defining the Target Probability

Calculating $P (X = 4)$

Calculating $P (X = 5)$

Combining the Probabilities

Simplifying the Final Fraction

Summary and Final Answer

The Sigma Insight: Binomial Distribution

The Architecture of Chance

Phase 1

Phase 2

Phase 3

Conclusion

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