The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability

Q: Wait, why is the variance $npq$? I thought variance was just the spread of data, why does it depend on $q$?

That is a brilliant question! In a binomial distribution, the variance measures the 'uncertainty' of the outcome. If $q$ (the probability of failure) is zero or one, there is no uncertainty—the outcome is deterministic. The $q$ term scales the variance based on how 'balanced' the success and failure probabilities are. It is the mathematical signature of the Bernoulli process, capturing exactly how much the data deviates from the mean!

Q: Can $n$ ever be a fraction? I got $n=8$, but what if I had calculated $n=7.5$?

In the context of a Binomial Distribution, $n$ represents the number of trials—it must be a non-negative integer. If you ever calculate a non-integer $n$, it is a clear signal to pause and re-check your arithmetic. You are counting discrete events, not measuring continuous quantities, so $n$ must always be a whole number.

Q: In the final step, why does the power of $1/2$ become $8$? I see $p^2$ and $q^6$, but how do they combine?

Great observation! Because $p = 1/2$ and $q = 1/2$, we have $(1/2)^2 \times (1/2)^6$. By the laws of exponents, when the bases are the same, we add the exponents: $2 + 6 = 8$. It is a beautiful simplification that happens because the probability of success and failure are equal in this specific problem. It makes the final calculation much cleaner!

Question

The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is

Options:

(A) $28/256$

(B) 219/256

(C) 128/256

(D) 37/256

Sigma Percentile · Accepted Answer

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

Welcome, student. Today, we are not just solving a problem; we are peeling back the layers of a Binomial Distribution to see how it functions.

Probability is often seen as a game of luck, but in the realm of JEE Advanced, it is a game of rigorous structure. We are given two vital statistics: the mean $\mu = 4$ and the variance $\sigma^2 = 2$. These two numbers are the DNA of our distribution.

### Phase 1: The Logic Bridge

In any binomial experiment, we have $n$ independent trials, each with a probability of success $p$ and a probability of failure $q$. We know the definitions by heart: the mean is $\mu = np$ and the variance is $\sigma^2 = npq$.

But how do we bridge the gap between these two? Look closely at the ratio of variance to mean:

$$
\frac{\sigma^2}{\mu} = \frac{npq}{np}
$$

The $n$ and $p$ terms cancel out beautifully, leaving us with just $q$. This is our 'Logic Bridge.' By dividing the variance by the mean, we isolate the probability of failure:

$$
q = \frac{2}{4} = \frac{1}{2}
$$

### Phase 2: Finding the Parameters

Now that we have $q = 1/2$, the rest of the puzzle falls into place with elegant simplicity. Since the sum of all probabilities in a binary outcome must be unity, we know that $p + q = 1$.

Therefore, $p = 1 - 1/2 = 1/2$. We have a perfectly balanced system where success and failure are equally likely.

With $p$ in hand, we return to our mean equation: $\mu = np = 4$. Substituting $p = 1/2$, we get:

$$
n 	imes \frac{1}{2} = 4
$$

Multiplying both sides by $2$, we find that the total number of trials is $n = 8$. We have successfully reconstructed the entire experiment from just two numbers!

### Phase 3: The Binomial Engine

Now, we engage the Binomial Probability formula: $P(X = r) = \binom{n}{r} p^r q^{n-r}$. This formula is the engine of our calculation.

We are tasked with finding the probability of exactly two successes, so we set $r = 2$. Substituting our known values, we get:

$$
P(X = 2) = \binom{8}{2} \left(\frac{1}{2}ight)^2 \left(\frac{1}{2}ight)^{8-2}
$$

Let us pause and appreciate the algebra here. We have $\left(\frac{1}{2}ight)^2$ multiplied by $\left(\frac{1}{2}ight)^6$. Because the bases are identical, we simply add the exponents: $2 + 6 = 8$.

This simplifies our expression to:

$$
P(X = 2) = \binom{8}{2} 	imes \left(\frac{1}{2}ight)^8
$$

### Phase 4: The Final Calculation

Finally, we calculate the combination $\binom{8}{2}$. This represents the number of ways to choose $2$ successes out of $8$ trials.

Using the formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$, we calculate:

$$
\binom{8}{2} = \frac{8 	imes 7}{2 	imes 1} = 28
$$

Now, we combine this with our power term: $P(X = 2) = 28 	imes \frac{1}{2^8}$. Since $2^8 = 256$, our final probability is:

$$
P(X = 2) = \frac{28}{256} = \mathbf{\frac{7}{64}}
$$

Look at the elegance of this result. We started with abstract parameters and arrived at a concrete probability. You have navigated the logic of binomial distributions with precision.

Animated Solution for Mathematics - Probability: The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is

Select Answer:

Visualized Solution

Identify Given Parameters

Relation Between Mean and Variance

Calculate Probability of Failure $q$

Calculate Probability of Success $p$

Calculate Number of Trials $n$

The Binomial Probability Formula

Substitute Values into the Formula

Calculate Combination $(2 8)$

Final Result and Conclusion

The Sigma Insight: Binomial Distribution

The Architecture of Chance

Phase 1

Phase 2

Phase 3

Phase 4

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