In a binomial distribution B(n, p), the sum and product of the mean & variance are 5 and 6 respectiv

Q: Wait, why do we reject the case where $q = 1.5$?

That is a fantastic question! In any binomial distribution, $q$ represents the probability of failure. By definition, any probability must lie in the interval $[0, 1]$. If you calculate $q = 1.5$, you have violated the fundamental axioms of probability. This is a classic JEE trap designed to test if you are blindly solving equations or thinking about the physical reality of the problem.

Q: Why can't we just use energy conservation or some other shortcut?

This isn't a physics problem, so energy conservation doesn't apply! However, you are right to look for shortcuts. The 'shortcut' here is the quadratic root property. By treating the mean and variance as roots of $t^2 - 5t + 6 = 0$, we bypass the need to solve a system of non-linear equations, which saves time and reduces the chance of algebraic errors.

Q: Is $n$ always an integer? I got $n = 9$, but what if I got a fraction?

Yes, $n$ must be a positive integer because it represents the number of trials in a binomial experiment. If you ever calculate a non-integer value for $n$, it is a clear signal that you have made a calculation error or chosen the wrong case for your mean and variance. Always double-check your arithmetic!

Question

In a binomial distribution $B(n, p)$, the sum and product of the mean & variance are 5 and 6 respectively, then find $6(n + p - q)$ is equal to

Options:

(A) 51

(B) 52

(C) 53

(D) 50

Sigma Percentile · Accepted Answer

### The Heartbeat of Uncertainty: Mastering the Binomial Distribution

Welcome, future engineer! Today, we are going to peel back the layers of a beautiful problem in probability. The binomial distribution, $B(n, p)$, is not just a collection of formulas; it is the mathematical heartbeat of uncertainty.

Whether you are predicting the outcome of a coin flip or the reliability of a circuit, this distribution is your best friend. Let us dive into this problem and see how we can turn a seemingly complex system of equations into a simple, elegant solution.

### Phase 1: The Foundation of Success and Failure

Every binomial experiment is defined by two parameters: $n$, the number of trials, and $p$, the probability of success. From these, we derive two critical metrics: the mean (or expected value) $\mu = np$ and the variance $\sigma^2 = npq$.

Here, $q = 1 - p$ is the probability of failure. Think of the mean as the 'center of gravity' of your outcomes and the variance as the 'spread' of those outcomes. The problem gives us two pieces of information: the sum of the mean and variance is $5$, and their product is $6$.

Mathematically, we have:

$$np + npq = 5$$

$$(np)(npq) = 6$$

I know what you are thinking—this looks like a system of non-linear equations. You could try to substitute $q = 1 - p$, but that leads to a messy cubic equation. Let us pause and look for a more elegant path.

### Phase 2: The Quadratic Trick

Whenever you see the sum and product of two quantities, your mind should immediately jump to the properties of quadratic equations. Let us define $X = np$ (the mean) and $Y = npq$ (the variance).

Our equations become:

$$X + Y = 5$$

$$XY = 6$$

Does this ring a bell? If $X$ and $Y$ are the roots of a quadratic equation in $t$, then the equation must be $t^2 - (X+Y)t + XY = 0$. Substituting our values, we get:

$$t^2 - 5t + 6 = 0$$

This is a standard, friendly quadratic. Factoring it, we get $(t - 2)(t - 3) = 0$, which gives us the roots $t = 2$ and $t = 3$. This means our mean and variance are either $2$ and $3$, or $3$ and $2$.

### Phase 3: The Logical Filter

This is the moment of truth. We have two cases:

**Case 1:** Mean $np = 2$, Variance $npq = 3$

**Case 2:** Mean $np = 3$, Variance $npq = 2$

To decide, we look at the ratio of variance to mean: $\frac{npq}{np} = q$. In Case 1, $q = \frac{3}{2} = 1.5$.

In Case 2, $q = \frac{2}{3} \approx 0.67$. Remember, $q$ is a probability, and probabilities cannot exceed $1$. Therefore, Case 1 is physically impossible. We must reject it. Case 2 is our winner: the mean is $3$ and the variance is $2$.

### Phase 4: Bringing It Home

Now that we have $q = \frac{2}{3}$, we find $p = 1 - q = \frac{1}{3}$. With $np = 3$, we substitute $p = \frac{1}{3}$ to find $n$: $n(\frac{1}{3}) = 3$, so $n = 9$.

We have all our variables: $n = 9$, $p = \frac{1}{3}$, and $q = \frac{2}{3}$. The final step is to evaluate the expression $6(n + p - q)$:

$$6\left(9 + \frac{1}{3} - \frac{2}{3}ight) = 6\left(9 - \frac{1}{3}ight)$$

Simplifying the fraction: $9 - \frac{1}{3} = \frac{27 - 1}{3} = \frac{26}{3}$.

Finally, $6 	imes \frac{26}{3} = 2 	imes 26 = 52$.

The **final answer is 52**. See how the complexity melted away? Keep practicing this logical approach, and you will master any problem JEE throws your way!

Animated Solution for Mathematics - Probability: In a binomial distribution $B (n, p)$ , the sum and product of the mean & variance are 5 and 6 respectively, then find $6 (n + p - q)$ is equal to

Select Answer:

Visualized Solution

Mean and Variance of $B (n, p)$

Setting up the Equations

Forming a Quadratic Equation

Solving the Quadratic Equation

Analyzing the Two Cases

Applying Probability Constraints

Calculating $p$ and $n$

Evaluating the Target Expression

Simplifying the Expression

Final Calculation

The Sigma Insight: Binomial Distribution

The Heartbeat of Uncertainty

Phase 1

Phase 2

Phase 3

Phase 4

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