The value of `P(2)` in a Binomial distribution when p=1/6 and n=5 is

Options

A. `(3125)/(7776)`
B. `(250)/(7776)`
C. `(1250)/(7776)`
D. `(25)/(7776)`

Answer

To find the value of \( P(2) \) in a Binomial distribution with \( p = \frac{1}{6} \) and \( n = 5 \), we will use the formula for the probability of getting exactly \( r \) successes in \( n \) trials:

\[ P(r) = \binom{n}{r} p^r q^{n-r} \]

where:
- \( \binom{n}{r} \) is the binomial coefficient, calculated as \( \frac{n!}{r!(n-r)!} \)
- \( p \) is the probability of success
- \( q = 1 - p \) is the probability of failure
- \( n \) is the number of trials
- \( r \) is the number of successes

### Step 1: Identify the parameters
Given:
- \( p = \frac{1}{6} \)
- \( n = 5 \)
- \( r = 2 \)

### Step 2: Calculate \( q \)
\[ q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6} \]

### Step 3: Calculate the binomial coefficient \( \binom{n}{r} \)
\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

### Step 4: Substitute the values into the formula
Now we can substitute \( n \), \( r \), \( p \), and \( q \) into the formula:
\[ P(2) = \binom{5}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{5-2} \]
\[ P(2) = 10 \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^3 \]

### Step 5: Calculate \( P(2) \)
Calculating each part:
\[ \left(\frac{1}{6}\right)^2 = \frac{1}{36} \]
\[ \left(\frac{5}{6}\right)^3 = \frac{125}{216} \]

Now substituting these values back:
\[ P(2) = 10 \times \frac{1}{36} \times \frac{125}{216} \]
\[ P(2) = 10 \times \frac{125}{7776} \]
\[ P(2) = \frac{1250}{7776} \]

### Step 6: Final Result
Thus, the value of \( P(2) \) is:
\[ P(2) = \frac{1250}{7776} \]