In a city 10 accident take place in a span of 50 days. Assuming that the number of accidents follow the Poisson distribution, the probability that three or more accident occure in a day , is

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the Poisson Distribution Parameter (λ)
Given that there are 10 accidents in 50 days, we can calculate the average number of accidents per day (λ).

\[ \lambda = \frac{\text{Total Accidents}}{\text{Total Days}} = \frac{10}{50} = 0.2 \]

### Step 2: Understand the Probability Requirement
We need to find the probability that 3 or more accidents occur in a day, which can be expressed as:

\[ P(X \geq 3) = 1 - P(X < 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2)) \]

### Step 3: Calculate Individual Probabilities
Using the Poisson probability mass function:

\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]

We will calculate \(P(X = 0)\), \(P(X = 1)\), and \(P(X = 2)\):

1. **For \(P(X = 0)\)**:
\[ P(X = 0) = \frac{e^{-0.2} (0.2)^0}{0!} = e^{-0.2} \cdot 1 = e^{-0.2} \]

2. **For \(P(X = 1)\)**:
\[ P(X = 1) = \frac{e^{-0.2} (0.2)^1}{1!} = e^{-0.2} \cdot 0.2 \]

3. **For \(P(X = 2)\)**:
\[ P(X = 2) = \frac{e^{-0.2} (0.2)^2}{2!} = e^{-0.2} \cdot \frac{0.04}{2} = e^{-0.2} \cdot 0.02 \]

### Step 4: Combine the Probabilities
Now we can add these probabilities together:

\[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \]

Substituting the values we calculated:

\[ P(X < 3) = e^{-0.2} + e^{-0.2} \cdot 0.2 + e^{-0.2} \cdot 0.02 \]
\[ = e^{-0.2} (1 + 0.2 + 0.02) = e^{-0.2} (1.22) \]

### Step 5: Calculate the Final Probability
Now we can find \(P(X \geq 3)\):

\[ P(X \geq 3) = 1 - P(X < 3) = 1 - e^{-0.2} \cdot 1.22 \]

### Final Answer
Thus, the probability that three or more accidents occur in a day is:

\[ P(X \geq 3) = 1 - 1.22 \cdot e^{-0.2} \]