Suppose the counts recorded by a Geiger counter follow a Poisson process with an average of two counts per minute. Round the answers to 3 decimal places.
(a) What is the probability that there are no counts in a 30-second interval?
(b) What is the probability that the first count occurs in less than 7 seconds?
(c) What is the probability that the first count occurs between 1 and 2 minutes after start-up?
To solve these questions, we need to use the properties of the Poisson distribution and the average rate of counts per minute. To find the answers, we can use the formula:
P(X = k) = (e^(-λ) * λ^k) / k!
Where:
P(X = k) is the probability of k counts occurring,
e is the base of the natural logarithm (approximately 2.71828),
λ is the average rate of counts per unit of time (in this case, per minute),
k is the number of counts.
Let's calculate the probabilities for each question:
(a) What is the probability that there are no counts in a 30-second interval?
To find the number of counts in a 30-second interval, we need to adjust the average rate of counts per minute to the equivalent rate per 30 seconds. Since there are 60 seconds in a minute, the rate per 30 seconds is λ/2.
Using k = 0 (since we want no counts) and λ = 2 (the average rate), we have:
P(X = 0) = (e^(-2/2) * (2/2)^0) / 0! = e^(-1)
Evaluating this expression, we find:
P(X = 0) = 0.367
Therefore, the probability that there are no counts in a 30-second interval is approximately 0.367.
(b) What is the probability that the first count occurs in less than 7 seconds?
To find this probability, we convert the given time to minutes and use the cumulative distribution function (CDF) of the Poisson distribution. The formula for the CDF is as follows:
P(X < k) = Σ(i=0 to k-1) [ (e^(-λ) * λ^i) / i! ]
In this case, we want to calculate P(X < 1). The average rate is still λ = 2 since it is defined per minute.
P(X < 1) = Σ(i=0 to 1-1) [ (e^(-2) * 2^i) / i! ] = (e^-2 * 2^0) / 0! = e^-2
Evaluating this expression, we find:
P(X < 1) = 0.135
Therefore, the probability that the first count occurs in less than 7 seconds is approximately 0.135.
(c) What is the probability that the first count occurs between 1 and 2 minutes after start-up?
To calculate this probability, we need to use the CDF again. This time, we find P(1 ≤ X ≤ 2). The average rate is still λ = 2.
P(1 ≤ X ≤ 2) = P(X = 1) + P(X = 2)
Using the formula for P(X = k) from earlier, we have:
P(X = 1) = (e^(-2) * 2^1) / 1! = 2 * e^-2
P(X = 2) = (e^(-2) * 2^2) / 2! = e^-2
Summing them up:
P(1 ≤ X ≤ 2) = 2 * e^-2 + e^-2 = 3 * e^-2
Evaluating this expression, we find:
P(1 ≤ X ≤ 2) = 0.406
Therefore, the probability that the first count occurs between 1 and 2 minutes after start-up is approximately 0.406.