If X has Poisson distribution with parameter m and `P(X = 2) = P(X = 3)`, then find `P(X ge 2).`
Use `e^(-3) = 0.0497`

Answer

To solve the problem, we need to find \( P(X \geq 2) \) given that \( P(X = 2) = P(X = 3) \) for a Poisson distribution with parameter \( m \).

### Step 1: Set up the equations for the Poisson probabilities

The probability mass function for a Poisson distribution is given by:

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

From the problem statement, we have:

\[ P(X = 2) = P(X = 3) \]

Substituting the formula for \( P(X = 2) \) and \( P(X = 3) \):

\[ \frac{e^{-m} m^2}{2!} = \frac{e^{-m} m^3}{3!} \]

### Step 2: Simplify the equation

We can cancel \( e^{-m} \) from both sides (assuming \( e^{-m} \neq 0 \)):

\[ \frac{m^2}{2!} = \frac{m^3}{3!} \]

Substituting the factorial values \( 2! = 2 \) and \( 3! = 6 \):

\[ \frac{m^2}{2} = \frac{m^3}{6} \]

### Step 3: Cross-multiply to eliminate the fractions

Cross-multiplying gives:

\[ 6m^2 = 2m^3 \]

### Step 4: Rearrange the equation

Rearranging the equation gives:

\[ 2m^3 - 6m^2 = 0 \]

Factoring out \( 2m^2 \):

\[ 2m^2(m - 3) = 0 \]

This gives us two solutions:

1. \( 2m^2 = 0 \) which implies \( m = 0 \) (not a valid solution for Poisson parameter)
2. \( m - 3 = 0 \) which implies \( m = 3 \)

### Step 5: Calculate \( P(X \geq 2) \)

Now that we have \( m = 3 \), we can find \( P(X \geq 2) \):

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

Calculating \( P(X = 0) \) and \( P(X = 1) \):

\[ P(X = 0) = \frac{e^{-3} 3^0}{0!} = e^{-3} \]
\[ P(X = 1) = \frac{e^{-3} 3^1}{1!} = 3e^{-3} \]

Thus,

\[ P(X < 2) = P(X = 0) + P(X = 1) = e^{-3} + 3e^{-3} = 4e^{-3} \]

Therefore,

\[ P(X \geq 2) = 1 - 4e^{-3} \]

### Step 6: Substitute the value of \( e^{-3} \)

Using the given value \( e^{-3} = 0.0497 \):

\[ P(X \geq 2) = 1 - 4(0.0497) = 1 - 0.1988 = 0.8012 \]

### Final Answer

Thus, the final result is:

\[ P(X \geq 2) = 0.8012 \]

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