For a binomial distribution B(n.p). np=4 and variance `npq=4//3`. What is the probability `P(x ge 5)` equal to?

Options

A. `(2//3)^(6)`
B. `2^(5)//3^(6)`
C. `(1//3)^(6)`
D. `2^(8)//3^(6)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given information
We are given:
- \( np = 4 \)
- Variance \( npq = \frac{4}{3} \)

### Step 2: Find \( q \)
From the variance formula, we know:
\[ npq = \frac{4}{3} \]
Substituting \( np = 4 \) into the variance equation:
\[ 4q = \frac{4}{3} \]
Now, solve for \( q \):
\[ q = \frac{4}{3} \div 4 = \frac{4}{3} \times \frac{1}{4} = \frac{1}{3} \]

### Step 3: Find \( p \)
Since \( p + q = 1 \):
\[ p = 1 - q = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 4: Find \( n \)
We know \( np = 4 \):
\[ n \cdot \frac{2}{3} = 4 \]
Now, solve for \( n \):
\[ n = 4 \div \frac{2}{3} = 4 \times \frac{3}{2} = 6 \]

### Step 5: Write the binomial distribution formula
The binomial distribution formula is given by:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]

### Step 6: Calculate \( P(X \geq 5) \)
We need to find \( P(X \geq 5) \), which is:
\[ P(X \geq 5) = P(X = 5) + P(X = 6) \]

#### Calculate \( P(X = 5) \)
Using the formula:
\[ P(X = 5) = \binom{6}{5} \left(\frac{2}{3}\right)^5 \left(\frac{1}{3}\right)^{6-5} \]
Calculating:
\[ \binom{6}{5} = 6 \]
Thus,
\[ P(X = 5) = 6 \left(\frac{2}{3}\right)^5 \left(\frac{1}{3}\right)^1 = 6 \cdot \frac{32}{243} \cdot \frac{1}{3} = \frac{192}{729} \]

#### Calculate \( P(X = 6) \)
Using the formula:
\[ P(X = 6) = \binom{6}{6} \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^{0} \]
Calculating:
\[ \binom{6}{6} = 1 \]
Thus,
\[ P(X = 6) = 1 \cdot \left(\frac{2}{3}\right)^6 = \frac{64}{729} \]

### Step 7: Combine the probabilities
Now, we combine both probabilities:
\[ P(X \geq 5) = P(X = 5) + P(X = 6) = \frac{192}{729} + \frac{64}{729} = \frac{256}{729} \]

### Final Answer
The probability \( P(X \geq 5) \) is:
\[ \frac{256}{729} \]