Suppose on an average 1 house in 1000 in a certain district has a fire during a year . If there are 2000 houses in the district , then the probability that exactly 5 houses will have a fire during the year is

Options

A. 1) `(1)/(15e^(2))`
B. 2) `(14)/(15e^(2))`
C. 3) `(4)/(15e^(2))`
D. 4) `e^(-2)`

Answer

To solve the problem, we will use the Poisson distribution formula. Let's go through the steps:

### Step 1: Determine the average rate (mean) of fires in the district.
Given that 1 house in 1000 has a fire on average, we can calculate the mean number of fires for 2000 houses.

\[ \text{Mean} (m) = \frac{2000 \text{ houses}}{1000} = 2 \]

### Step 2: Identify the value of \(x\).
We need to find the probability that exactly 5 houses will have a fire during the year. Thus, we have:

\[ x = 5 \]

### Step 3: Write the Poisson probability formula.
The Poisson probability formula is given by:

\[ P(X = x) = \frac{e^{-m} \cdot m^x}{x!} \]

### Step 4: Substitute the values into the formula.
Now, we substitute \(m = 2\) and \(x = 5\) into the formula:

\[ P(X = 5) = \frac{e^{-2} \cdot 2^5}{5!} \]

### Step 5: Calculate \(2^5\) and \(5!\).
Calculating these values:

\[ 2^5 = 32 \]

\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]

### Step 6: Substitute these values back into the formula.
Now, substituting these values back into the probability formula:

\[ P(X = 5) = \frac{e^{-2} \cdot 32}{120} \]

### Step 7: Simplify the expression.
We can simplify this expression:

\[ P(X = 5) = \frac{32}{120} \cdot e^{-2} = \frac{8}{30} \cdot e^{-2} = \frac{4}{15} \cdot e^{-2} \]

### Final Answer:
Thus, the probability that exactly 5 houses will have a fire during the year is:

\[ P(X = 5) = \frac{4}{15} e^{-2} \]

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