Determine the binomial distribution where means 9 and standard deviation is `3/(2)`. Also, find the probability of obtainning at most one success.

Answer

To determine the binomial distribution with a mean of 9 and a standard deviation of \( \frac{3}{2} \), and to find the probability of obtaining at most one success, we can 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.

Given that the mean is 9:
\[ N \cdot P = 9 \tag{1} \]

The standard deviation \( \sigma \) of a binomial distribution is given by:
\[ \sigma = \sqrt{N \cdot P \cdot Q} \]
where \( Q = 1 - P \).

Given that the standard deviation is \( \frac{3}{2} \):
\[ \sqrt{N \cdot P \cdot Q} = \frac{3}{2} \]
Squaring both sides gives:
\[ N \cdot P \cdot Q = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \tag{2} \]

### Step 2: Substitute \( Q \) in terms of \( P \)
From equation (2), we know:
\[ Q = 1 - P \]
Substituting this into equation (2):
\[ N \cdot P \cdot (1 - P) = \frac{9}{4} \]

### Step 3: Solve for \( N \) and \( P \)
Now we have two equations:
1. \( N \cdot P = 9 \)
2. \( N \cdot P \cdot (1 - P) = \frac{9}{4} \)

Substituting \( N \cdot P = 9 \) into the second equation:
\[ 9 \cdot (1 - P) = \frac{9}{4} \]
Dividing both sides by 9:
\[ 1 - P = \frac{1}{4} \]
Thus:
\[ P = 1 - \frac{1}{4} = \frac{3}{4} \]

### Step 4: Find \( N \)
Now substitute \( P \) back into equation (1):
\[ N \cdot \frac{3}{4} = 9 \]
Solving for \( N \):
\[ N = 9 \cdot \frac{4}{3} = 12 \]

### Step 5: Summary of parameters
We have determined:
- \( N = 12 \)
- \( P = \frac{3}{4} \)
- \( Q = 1 - P = \frac{1}{4} \)

### Step 6: Find the probability of obtaining at most one success
The probability of obtaining at most one success is given by:
\[ P(X \leq 1) = P(X = 0) + P(X = 1) \]

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

Calculating \( P(X = 0) \):
\[ P(X = 0) = \binom{12}{0} \left(\frac{3}{4}\right)^0 \left(\frac{1}{4}\right)^{12} = 1 \cdot 1 \cdot \left(\frac{1}{4}\right)^{12} = \left(\frac{1}{4}\right)^{12} \]

Calculating \( P(X = 1) \):
\[ P(X = 1) = \binom{12}{1} \left(\frac{3}{4}\right)^1 \left(\frac{1}{4}\right)^{11} = 12 \cdot \frac{3}{4} \cdot \left(\frac{1}{4}\right)^{11} = 12 \cdot \frac{3}{4} \cdot \frac{1}{4^{11}} = 12 \cdot \frac{3}{4^{12}} = \frac{36}{4^{12}} \]

### Step 7: Combine the probabilities
Now combine the two probabilities:
\[ P(X \leq 1) = P(X = 0) + P(X = 1) = \left(\frac{1}{4}\right)^{12} + \frac{36}{4^{12}} = \frac{1 + 36}{4^{12}} = \frac{37}{4^{12}} \]

### Final Answer
Thus, the probability of obtaining at most one success is:
\[ P(X \leq 1) = \frac{37}{4^{12}} \]