A random variable X follows binomial probability distribution with probability P(X), with mean as 2, probability of success as p and probability of failure as q such that `p+q=1.` If `SigmaX^(2)P(X)=(28)/(5)`, then the probability of exactly 2 success is

Options

A. `(3xx2^(14))/(5^(10))`
B. `(3^(2)xx2^(18))/(5^(9))`
C. `3xx((2)/(5))^(10)`
D. `45xx((2)/(5))^(9)`

Answer

To solve the problem step by step, we will follow the details provided in the video transcript.

### Step 1: Understand the Mean of the Binomial Distribution
The mean of a binomial distribution is given by the formula:
\[ \text{Mean} = Np \]
where \(N\) is the number of trials and \(p\) is the probability of success. We are given that the mean is 2:
\[ Np = 2 \]

### Step 2: Understand the Variance of the Binomial Distribution
The variance of a binomial distribution is given by:
\[ \text{Variance} = Npq \]
where \(q = 1 - p\). We are also given that:
\[ \Sigma X^2 P(X) = \frac{28}{5} \]
Using the formula for variance, we can express it as:
\[ Npq = \Sigma X^2 P(X) - \text{Mean}^2 \]
Substituting the known values:
\[ Npq = \frac{28}{5} - 2^2 = \frac{28}{5} - 4 = \frac{28}{5} - \frac{20}{5} = \frac{8}{5} \]

### Step 3: Relate \(N\), \(p\), and \(q\)
From the mean, we have:
\[ Np = 2 \]
From the variance, we have:
\[ Npq = \frac{8}{5} \]
Substituting \(q = 1 - p\) into the variance equation:
\[ Np(1 - p) = \frac{8}{5} \]
Substituting \(Np = 2\):
\[ 2(1 - p) = \frac{8}{5} \]
Solving for \(p\):
\[ 1 - p = \frac{4}{5} \implies p = 1 - \frac{4}{5} = \frac{1}{5} \]

### Step 4: Find \(q\)
Now, since \(q = 1 - p\):
\[ q = 1 - \frac{1}{5} = \frac{4}{5} \]

### Step 5: Find \(N\)
Using the mean equation:
\[ Np = 2 \implies N \cdot \frac{1}{5} = 2 \implies N = 2 \cdot 5 = 10 \]

### Step 6: Calculate the Probability of Exactly 2 Successes
The probability of getting exactly \(r\) successes in a binomial distribution is given by:
\[ P(X = r) = \binom{N}{r} p^r q^{N-r} \]
For \(r = 2\):
\[ P(X = 2) = \binom{10}{2} \left(\frac{1}{5}\right)^2 \left(\frac{4}{5}\right)^{10-2} \]
Calculating \( \binom{10}{2} \):
\[ \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \]
Now substituting into the probability formula:
\[ P(X = 2) = 45 \left(\frac{1}{5}\right)^2 \left(\frac{4}{5}\right)^8 \]
Calculating:
\[ = 45 \cdot \frac{1}{25} \cdot \left(\frac{4^8}{5^8}\right) = 45 \cdot \frac{1}{25} \cdot \frac{65536}{390625} \]
Simplifying:
\[ = \frac{45 \cdot 65536}{25 \cdot 390625} \]
Calculating the final result:
\[ = \frac{2949120}{9765625} \]

### Final Answer
Thus, the probability of exactly 2 successes is:
\[ P(X = 2) = \frac{2949120}{9765625} \]