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.
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 since the average rate of rescue calls per day is given.
The Poisson distribution formula is:
P(x; λ) = (e^(-λ) * λ^x) / x!
Where:
- P(x; λ) is the probability of getting exactly x events when the average rate is λ
- e is Euler's number (approximately 2.71828)
- λ is the average rate of events per interval (in this case, per day)
- x is the actual number of events
In this case, λ = 0.5 (average number of rescue calls per day) and we need to find the probability of getting fewer than two calls, so x = 0 and x = 1.
Let's calculate the probability step by step:
P(x = 0) = (e^(-0.5) * 0.5^0) / 0! = e^(-0.5) = 0.60653 (approximately)
P(x = 1) = (e^(-0.5) * 0.5^1) / 1! = 0.5 * e^(-0.5) = 0.30326 (approximately)
Now, to find the probability of receiving fewer than two calls, we sum the probabilities of receiving zero and one call:
P(X < 2) = P(x = 0) + P(x = 1) = 0.60653 + 0.30326 = 0.90979 (approximately)
Therefore, the probability that on a randomly selected day the fire station will receive fewer than two calls is approximately 0.90979, or about 90.98%.