Question 1184195
Here's how to solve this problem using the Poisson distribution:

**a. Probability of Exactly Three Suicides:**

The Poisson probability formula is:

P(x) = (e^-μ * μ^x) / x!

Where:

* x is the number of events (suicides in this case)
* μ is the average number of events (2.75 in this case)
* e is the base of the natural logarithm (~2.71828)
* x! is the factorial of x

We want to find P(3), the probability of exactly three suicides:

P(3) = (e^-2.75 * 2.75^3) / 3!
P(3) = (0.0639 * 20.796875) / 6
P(3) = 1.3307 / 6
P(3) ≈ 0.2218

So, the probability that a randomly selected month will have exactly three suicides is approximately 0.2218 or 22.18%.

**b. Probability of At Least Three Suicides:**

To find the probability of at least three suicides, we need to calculate P(x ≥ 3).  This is equal to 1 minus the probability of zero, one, or two suicides:

P(x ≥ 3) = 1 - [P(0) + P(1) + P(2)]

Let's calculate P(0), P(1), and P(2):

* P(0) = (e^-2.75 * 2.75^0) / 0! = (0.0639 * 1) / 1 ≈ 0.0639
* P(1) = (e^-2.75 * 2.75^1) / 1! = (0.0639 * 2.75) / 1 ≈ 0.1757
* P(2) = (e^-2.75 * 2.75^2) / 2! = (0.0639 * 7.5625) / 2 ≈ 0.2415

Now, substitute these values back into the equation for P(x ≥ 3):

P(x ≥ 3) = 1 - (0.0639 + 0.1757 + 0.2415)
P(x ≥ 3) = 1 - 0.4811
P(x ≥ 3) ≈ 0.5189

Therefore, the probability that a randomly selected month will have at least three suicides is approximately 0.5189 or 51.89%.