A local fire station receives an average of 0.5 rescue calls per day. Find the probability that on a randomly selected day, the fire station will receive fewer than two calls. (Round your answer to the nearest thousandth.)

To find the probability that the fire station will receive fewer than two calls on a randomly selected day, we can use the concept of a Poisson distribution.

The Poisson distribution is used to model the number of events that occur in a fixed interval of time or space, given the average rate of occurrence. In this case, the average rate of rescue calls is 0.5 per day.

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

P(X=k) = (e^(-λ) * λ^k) / k!

Where X is the random variable representing the number of events, λ is the average rate of occurrence, e is the base of the natural logarithm, and k is the actual number of events.

Since we want to calculate the probability of fewer than two calls, we need to calculate the probabilities of getting zero and one calls, and then sum them.

P(X<2) = P(X=0) + P(X=1)

Let's calculate each individual probability:

P(X=0) = (e^(-0.5) * (0.5^0)) / 0! = e^(-0.5)

e is approximately equal to 2.71828, so we can use this value to calculate e^(-0.5).

P(X=0) ≈ 2.71828^(-0.5) ≈ 0.60653

P(X=1) = (e^(-0.5) * (0.5^1)) / 1! = (e^(-0.5) * 0.5) = 0.5 * e^(-0.5)

Now we can add these probabilities together to find the final answer:

P(X<2) ≈ 0.60653 + 0.5 * e^(-0.5)

Using a calculator, we can get a numerical value for this sum:

P(X<2) ≈ 0.60653 + 0.5 * 0.60653 ≈ 0.60653 + 0.303265 ≈ 0.909795

Therefore, the probability that on a randomly selected day the fire station will receive fewer than two calls is approximately 0.910 (rounded to the nearest thousandth).

To find the probability that the fire station will receive fewer than two calls on a randomly selected day, we can use the Poisson distribution.

The formula for the Poisson probability is:

P(x; μ) = e^(-μ) * (μ^x) / x!

Where:
- P(x; μ) is the probability of getting exactly x successes,
- e is the base of the natural logarithm (approximately 2.71828),
- μ is the average number of successes in a given time interval,
- x is the number of successes.

Here, μ = 0.5 (average number of rescue calls per day) and we need to find P(x < 2).

P(x < 2) = P(x = 0) + P(x = 1)

Let's calculate each part step by step:

For x = 0:
P(x = 0) = e^(-μ) * (μ^0) / 0!
= e^(-0.5) * (0.5^0) / 0!
= e^(-0.5) / 1
≈ 0.606

For x = 1:
P(x = 1) = e^(-μ) * (μ^1) / 1!
= e^(-0.5) * (0.5^1) / 1!
= e^(-0.5) * 0.5 / 1
≈ 0.303

Therefore, P(x < 2) = P(x = 0) + P(x = 1)
≈ 0.606 + 0.303
≈ 0.909

So, the probability that on a randomly selected day, the fire station will receive fewer than two calls is approximately 0.909.