Suppose that the number of patients admitted into a hospital emergency room during a certain hour of the day has a Poisson distribution with mean 3. Find the probability that durin... Suppose that the number of patients admitted into a hospital emergency room during a certain hour of the day has a Poisson distribution with mean 3. Find the probability that during that hour there will be exactly two emergency patients. Assume that e^(-3) = 0.05. Read more

Steps to Solve

1. Recall the Poisson Probability Formula

   The Poisson probability formula is given by:

   $P(x) = \frac{\lambda^x e^{-\lambda}}{x!}$

   where:

   • $P(x)$ is the probability of $x$ events occurring.
   • $\lambda$ is the average rate of events (in this case, the mean number of patients admitted).
   • $e$ is the base of the natural logarithm (approximately 2.71828).
   • $x!$ is the factorial of $x$.

2. Identify Given Values

   From the problem statement, we have:

   • $\lambda = 3$ (the mean number of patients)
   • $x = 2$ (we want to find the probability of exactly 2 patients)
   • $e^{-3} = 0.05$ (given)

3. Apply the Formula

   Plug the values into the Poisson formula:

   $P(2) = \frac{3^2 \cdot e^{-3}}{2!}$

4. Calculate the Probability

   Substitute the given value of $e^{-3}$ and simplify:

   $P(2) = \frac{3^2 \cdot 0.05}{2!}$

   $P(2) = \frac{9 \cdot 0.05}{2}$

   $P(2) = \frac{0.45}{2}$

   $P(2) = 0.225$