Questions: Given that x has a Poisson distribution with μ=10, what is the probability that x=3? P(3) ≈ (Round to four decimal places as needed.)

Solution Steps

The probability of observing exactly \( k \) events in a Poisson distribution is given by the formula:

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

where:

• \( \lambda \) is the average number of events (rate parameter),
• \( k \) is the number of events for which the probability is calculated,
• \( e \) is the base of the natural logarithm.

For this problem, we have:

• \( \lambda = 10 \)
• \( k = 3 \)

Substituting these values into the formula, we get:

\[ P(X = 3) = \frac{10^3 e^{-10}}{3!} \]

Calculating each component:

• \( 10^3 = 1000 \)
• \( 3! = 6 \)
• \( e^{-10} \approx 0.0000453999 \) (retaining the original decimal notation for precision)

Now substituting these values back into the equation:

\[ P(X = 3) = \frac{1000 \cdot 0.0000453999}{6} \]

Calculating the final probability:

\[ P(X = 3) \approx \frac{0.0453999}{6} \approx 0.00756665 \]

Rounding to four decimal places, we find:

\[ P(X = 3) \approx 0.0076 \]

Final Answer