Question

we place at random 200 points in the interval (0, 100). Find the probability that in the interval (0, 2) there will be one and only one print, (compute exactly at first and then using the poisson approximation)

Answer 1

The probability of having one and only one point in the interval (0, 2) is approximately 0.0733.

Step-by-step explanation:

To find the probability that in the interval (0, 2) there will be one and only one point, we can use the concept of the Poisson distribution.

Let's calculate it exactly first:

The average rate of points in the interval (0, 100) is given by (200 points) / (100 units) = 2 points per unit.

In the interval (0, 2), the expected number of points is 2 points per unit * 2 units = 4 points.

Using the Poisson probability formula, the probability of having exactly one point in the interval (0, 2) is:

P(X = 1) = (4^1 * e^(-4)) / 1! ≈ 0.0733

Now let's use the Poisson approximation.

Since the interval (0, 2) is small and the expected number of points is low, we can approximate it using the Poisson distribution.

The parameter λ (lambda) for the Poisson distribution is equal to the average rate of points in the interval (0, 2), which is 2 points per unit * 2 units = 4.

Using the Poisson probability formula with λ = 4 and X = 1, the approximation gives us:

P(X = 1) = (4^1 * e^(-4)) / 1! ≈ 0.0733

Therefore the probability is 0.0733