If X follows a binomial distribution with mean 4 and variance 2, find `P(X ge5)`

Answer

To solve the problem where \( X \) follows a binomial distribution with a mean of 4 and a variance of 2, we will follow these steps:

### Step 1: Identify the parameters of the binomial distribution
The mean \( \mu \) of a binomial distribution is given by:
\[ \mu = n \cdot p \]
where \( n \) is the number of trials and \( p \) is the probability of success.

The variance \( \sigma^2 \) is given by:
\[ \sigma^2 = n \cdot p \cdot (1 - p) \]
where \( 1 - p \) is the probability of failure.

### Step 2: Set up the equations based on the given mean and variance
From the problem, we have:
\[ n \cdot p = 4 \quad \text{(1)} \]
\[ n \cdot p \cdot (1 - p) = 2 \quad \text{(2)} \]

### Step 3: Substitute equation (1) into equation (2)
From equation (1), we can express \( n \) in terms of \( p \):
\[ n = \frac{4}{p} \]
Now substitute \( n \) into equation (2):
\[ \frac{4}{p} \cdot p \cdot (1 - p) = 2 \]
This simplifies to:
\[ 4(1 - p) = 2 \]

### Step 4: Solve for \( p \)
Expanding the equation gives:
\[ 4 - 4p = 2 \]
Rearranging gives:
\[ 4p = 2 \implies p = \frac{1}{2} \]

### Step 5: Find \( n \)
Substituting \( p \) back into equation (1):
\[ n \cdot \frac{1}{2} = 4 \implies n = 8 \]

### Step 6: Calculate \( q \)
Since \( q = 1 - p \):
\[ q = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 7: Find \( P(X \geq 5) \)
We need to calculate:
\[ P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) \]
Using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
Substituting \( n = 8 \), \( p = \frac{1}{2} \), and \( q = \frac{1}{2} \):
\[ P(X = k) = \binom{8}{k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{8-k} = \binom{8}{k} \left(\frac{1}{2}\right)^8 \]

### Step 8: Calculate each probability
Now we calculate:
\[ P(X = 5) = \binom{8}{5} \left(\frac{1}{2}\right)^8 \]
\[ P(X = 6) = \binom{8}{6} \left(\frac{1}{2}\right)^8 \]
\[ P(X = 7) = \binom{8}{7} \left(\frac{1}{2}\right)^8 \]
\[ P(X = 8) = \binom{8}{8} \left(\frac{1}{2}\right)^8 \]

Calculating the binomial coefficients:
\[ \binom{8}{5} = 56, \quad \binom{8}{6} = 28, \quad \binom{8}{7} = 8, \quad \binom{8}{8} = 1 \]

Thus:
\[ P(X = 5) = \frac{56}{256}, \quad P(X = 6) = \frac{28}{256}, \quad P(X = 7) = \frac{8}{256}, \quad P(X = 8) = \frac{1}{256} \]

### Step 9: Sum the probabilities
Now, sum these probabilities:
\[ P(X \geq 5) = \frac{56 + 28 + 8 + 1}{256} = \frac{93}{256} \]

### Final Answer
Thus, the probability \( P(X \geq 5) \) is:
\[ \boxed{\frac{93}{256}} \]