Suppose we want to model the number of rats captured in a single trap in a day, where the average
number caught is 5 (i.e. X ∼ Poi(5)).
(a) What is the probability of catching 6 rats in one trap in a day?
(b) What is the probability of catching 12 rats in 2 traps in a single day?
To find the probabilities in these scenarios, we need to use the probability mass function (PMF) of the Poisson distribution, which is given by P(X = k) = (e^(-λ) * λ^k) / k!, where X is the random variable, λ is the average number of events, and k is the number of events.
(a) What is the probability of catching 6 rats in one trap in a day?
To find this probability, we can plug in the given values into the PMF formula:
P(X = 6) = (e^(-5) * 5^6) / 6!
To calculate this value, follow these steps:
1. Calculate e^(-5): Raise the constant e to the power of -5. This is approximately 0.0067379.
2. Calculate 5^6: Raise 5 to the power of 6. This equals 15625.
3. Calculate 6!: Calculate the factorial of 6 (6! = 6 * 5 * 4 * 3 * 2 * 1 = 720).
Now, plug these values into the formula:
P(X = 6) = (0.0067379 * 15625) / 720
Calculating this expression will give you the probability of catching 6 rats in one trap in a day.
(b) What is the probability of catching 12 rats in 2 traps in a single day?
To find this probability, we need to consider the average number of events happening in each trap, which is still 5, but now we have 2 traps. So the average number of events for 2 traps is 5 * 2 = 10.
Now, we can use the same PMF formula but with λ = 10 and k = 12:
P(X = 12) = (e^(-10) * 10^12) / 12!
Follow the same steps as in part (a) to calculate this probability.