Let X have a binomial distribution B(n, p) such that the sum and the product of the mean and varianc

Q: Why can't we just use energy conservation?

This is a probability problem, not a mechanics problem! Energy conservation doesn't apply here. We use the properties of the binomial distribution, specifically the definitions of mean and variance.

Q: Why do we reject the negative root?

In this problem, the roots are both positive ($16$ and $8$). If you ever encounter a negative root in a probability problem, you must reject it because mean and variance (for a binomial distribution) must be non-negative.

Q: Does $P(X > n - 3)$ include $n - 3$?

No, the inequality is strict. $X > n - 3$ means $X$ must be $n - 2, n - 1,$ or $n$. Since $n = 32$, we are looking for $P(X > 29)$, which is $P(X=30) + P(X=31) + P(X=32)$.

Question

Let $X$ have a binomial distribution $B(n, p)$ such that the sum and the product of the mean and variance of $X$ are 24 and 128 respectively. If $P(X > n - 3) = \frac{k}{2^n}$, then $k$ is equal to

Options:

(A) 528

(B) 529

(C) 629

(D) 630

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the threshold of a classic probability problem. You have a binomial distribution $B(n, p)$, and you are given two pieces of information: the sum of its mean and variance is $24$, and their product is $128$.

For a binomial distribution, the mean is $\mu = np$ and the variance is $\sigma^2 = npq$, where $q = 1 - p$. We are given the following system of equations:

$$np + npq = 24$$

$$(np)(npq) = 128$$

### Decoding the Mean and Variance

This is a classic setup for a quadratic equation. If we let $t$ be a variable representing these two quantities, then the quadratic equation is $t^2 - (	ext{Sum})t + (	ext{Product}) = 0$.

Substituting our values, we get:

$$t^2 - 24t + 128 = 0$$

Solving this quadratic equation, we find the roots are $16$ and $8$.

Now, here is the crucial physical reality check: since $q \le 1$, it must be that $np \ge npq$. Thus, the mean must be $np = 16$ and the variance must be $npq = 8$.

### Finding the Parameters

With $np = 16$ and $npq = 8$, we can easily find $q$ by taking the ratio:

$$\frac{npq}{np} = \frac{8}{16} = \frac{1}{2}$$

Since $q = \frac{1}{2}$, it follows that $p = 1 - q = \frac{1}{2}$. Now, substituting $p = \frac{1}{2}$ into $np = 16$, we find:

$$n \left( \frac{1}{2} ight) = 16 \Rightarrow n = 32$$

We have successfully decoded the entire distribution parameters as $n = 32$ and $p = \frac{1}{2}$.

### The Final Calculation

The question asks for $P(X > n - 3)$. With $n = 32$, this is $P(X > 29)$, which is equivalent to:

$$P(X = 30) + P(X = 31) + P(X = 32)$$

Using the binomial formula $P(X=r) = {}^{n}C_{r} p^r q^{n-r}$, and noting that $p = q = \frac{1}{2}$, the probability becomes:

$$P(X > 29) = \frac{{}^{32}C_{30} + {}^{32}C_{31} + {}^{32}C_{32}}{2^{32}}$$

Calculating the combinations:

$${}^{32}C_{30} = {}^{32}C_{2} = \frac{32 	imes 31}{2} = 496$$

$${}^{32}C_{31} = 32$$

$${}^{32}C_{32} = 1$$

The sum is $496 + 32 + 1 = 529$. Thus, the final probability is:

$$P(X > 29) = \frac{529}{2^{32}}$$

Comparing this to the form $\frac{k}{2^n}$, we find **$k = 529$**.

Animated Solution for Mathematics - Probability: Let $X$ have a binomial distribution $B (n, p)$ such that the sum and the product of the mean and variance of $X$ are 24 and 128 respectively. If $P (X > n - 3) = \frac{k}{2 ^{n}}$ , then $k$ is equal to

Select Answer:

Visualized Solution

Define Mean and Variance

Set up Equations

Form the Quadratic Equation

Solve for Roots

Identify Mean and Variance

Solve for p and q

Find the value of n

Define Target Probability

Apply Binomial Formula

Calculate Combinations

Find the Final Sum and k

The Sigma Insight: Binomial Distribution

Analyzing the Setup

Decoding the Mean and Variance

Finding the Parameters

The Final Calculation

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