The mean and variance of a Binomial variable X are respectively 4 and `(4)/(5)`. Find `P(X ge 3)`.

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the parameters of the Binomial distribution
The mean (μ) and variance (σ²) of a Binomial distribution are given by:
- Mean: \( \mu = np \)
- Variance: \( \sigma^2 = npq \)

Where:
- \( n \) = number of trials
- \( p \) = probability of success
- \( q \) = probability of failure = \( 1 - p \)

From the problem, we have:
- \( np = 4 \) (Equation 1)
- \( npq = \frac{4}{5} \) (Equation 2)

### Step 2: Express \( q \) in terms of \( p \)
From Equation 2, we can express \( q \):
\[ q = 1 - p \]
Substituting this into Equation 2 gives:
\[ np(1 - p) = \frac{4}{5} \]

### Step 3: Substitute \( np \) from Equation 1
Substituting \( np = 4 \) into the equation:
\[ 4(1 - p) = \frac{4}{5} \]

### Step 4: Solve for \( p \)
Dividing both sides by 4:
\[ 1 - p = \frac{1}{5} \]
Thus:
\[ p = 1 - \frac{1}{5} = \frac{4}{5} \]

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

### Step 6: Substitute \( p \) into Equation 1 to find \( n \)
Using \( np = 4 \):
\[ n \cdot \frac{4}{5} = 4 \]
Solving for \( n \):
\[ n = 4 \cdot \frac{5}{4} = 5 \]

### Step 7: Calculate \( P(X \geq 3) \)
To find \( P(X \geq 3) \), we calculate:
\[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) \]

Using the binomial probability formula:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]

Calculating each term:
1. **For \( P(X = 3) \)**:
\[ P(X = 3) = \binom{5}{3} \left(\frac{4}{5}\right)^3 \left(\frac{1}{5}\right)^{2} \]
Calculating:
\[ = 10 \cdot \left(\frac{64}{125}\right) \cdot \left(\frac{1}{25}\right) = \frac{640}{3125} \]

2. **For \( P(X = 4) \)**:
\[ P(X = 4) = \binom{5}{4} \left(\frac{4}{5}\right)^4 \left(\frac{1}{5}\right)^{1} \]
Calculating:
\[ = 5 \cdot \left(\frac{256}{625}\right) \cdot \left(\frac{1}{5}\right) = \frac{256}{3125} \]

3. **For \( P(X = 5) \)**:
\[ P(X = 5) = \binom{5}{5} \left(\frac{4}{5}\right)^5 \left(\frac{1}{5}\right)^{0} \]
Calculating:
\[ = 1 \cdot \left(\frac{1024}{3125}\right) = \frac{1024}{3125} \]

### Step 8: Sum the probabilities
Now we sum these probabilities:
\[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) \]
\[ = \frac{640}{3125} + \frac{256}{3125} + \frac{1024}{3125} \]
\[ = \frac{640 + 256 + 1024}{3125} = \frac{1920}{3125} \]

### Final Answer
Thus, the probability \( P(X \geq 3) \) is:
\[ \boxed{\frac{1920}{3125}} \]