Question

If \(X\sim(5,p)\) \(P(X=3) = 5P(X=4)\), find variance.

Answer 1

1. Understand the Problem:
We're given that X follows a binomial distribution with n = 5 trials and probability of success p, denoted as X ~ B(5, p). We're also given that P(X = 3) = 5P(X = 4). We need to find the variance of X.

2. Recall the Binomial Probability Formula:
The probability of getting exactly k successes in n trials is given by:
P(X = k) = (nCk) * p^k * (1 - p)^(n - k)
where (nCk) = n! / (k! * (n - k)!)

3. Set Up the Given Equation:
We are given P(X = 3) = 5P(X = 4). Let's write out the probabilities using the binomial probability formula:

P(X = 3) = (5C3) * p^3 * (1 - p)^(5 - 3) = (5C3) * p^3 * (1 - p)^2
P(X = 4) = (5C4) * p^4 * (1 - p)^(5 - 4) = (5C4) * p^4 * (1 - p)

4. Substitute and Solve for p:
Now substitute these into the given equation:

(5C3) * p^3 * (1 - p)^2 = 5 * (5C4) * p^4 * (1 - p)

Calculate the binomial coefficients:

(5C3) = 5! / (3! * 2!) = (5 * 4) / 2 = 10
(5C4) = 5! / (4! * 1!) = 5

Substitute these values back into the equation:

10 * p^3 * (1 - p)^2 = 5 * 5 * p^4 * (1 - p)
10 * p^3 * (1 - p)^2 = 25 * p^4 * (1 - p)

Now, we can simplify:

* Divide both sides by 5: 2 * p^3 * (1 - p)^2 = 5 * p^4 * (1 - p)
* Divide both sides by p^3 (assuming p ≠ 0): 2 * (1 - p)^2 = 5 * p * (1 - p)
* Divide both sides by (1 - p) (assuming p ≠ 1): 2 * (1 - p) = 5 * p
* 2 - 2p = 5p
* 2 = 7p
* p = 2/7

5. Calculate the Variance:
The variance of a binomial distribution is given by:

Variance = np(1 - p)

Substitute n = 5 and p = 2/7:

Variance = 5 * (2/7) * (1 - 2/7)
Variance = 5 * (2/7) * (5/7)
Variance = 50 / 49

Therefore, the variance is 50/49.

Answer 2

Given:
\(X \sim \text{Binomial}(5, p)\)
\(P(X=3) = 5P(X=4)\)

Binomial probabilities:
Probability mass function (pmf) for X:
\(P(X=k) = \binom{5}{k} p^k (1-p)^{5-k}\)

Calculate \( P(X=3) \) and \( P(X=4) \):
\(P(X=3) = \binom{5}{3} p^3 (1-p)^{2} = 10 p^3 (1-p)^{2}\)

\(P(X=4) = \binom{5}{4} p^4 (1-p)^{1} = 5 p^4 (1-p)\)

\(P(X=3) = 5 P(X=4)\)
\(10 p^3 (1-p)^{2} = 5 \cdot 5 p^4 (1-p)\)
\(10 p^3 (1-p)^{2} = 25 p^4 (1-p)\)
Divide both sides by \(p^3\) (assuming \(p \neq 0\)):
\(10 (1-p)^{2} = 25 p (1-p)\)
Divide both sides by \((1-p)\) (assuming \(p \neq 1\)):
\(10 (1-p) = 25 p\)
\(10 - 10p = 25p\)
\(10 = 35p\)
\(p = \frac{10}{35} = \frac{2}{7}\)

Calculate the variance of X:
Variance of a binomial random variable \(X \sim \text{Binomial}(n, p)\) is \(\text{Var}(X) = n p (1-p)\).
Substitute \(n = 5\) and \(p = \frac{2}{7}\):
\(\text{Var}(X) = 5 \cdot \frac{2}{7} \cdot \left( 1 - \frac{2}{7} \right)\)

\(\text{Var}(X) = 5 \cdot \frac{2}{7} \cdot \frac{5}{7}\)

\(\text{Var}(X) = \frac{50}{49}\)

So, the answer is: \(\frac{50}{49}\)