Ten participants are randomly selected from a population of patients with history of heart disease to be tracked in a health study. The study begins with an initial assessment of health status. The participants are instructed to return for a follow up visit one year after the initial assessment.
For these ten participants, the time (in years) from the initial assessment to the first heart attack has an exponential distribution with mean 5.
The times to first heart attack for these participants are independent.
What is the probability that prior to the one-year follow up visit, three or more of the participants have experienced their first heart attack since initial assessment?
To find the probability that three or more participants have experienced their first heart attack prior to the one-year follow-up visit, we can use the exponential distribution and the concept of the cumulative distribution function (CDF).
The exponential distribution with a mean of 5 years has the following cumulative distribution function:
CDF(x) = 1 - e^(-x/5)
To find the probability that a randomly selected participant has not experienced their first heart attack before the one-year follow-up visit (which is equivalent to not having a heart attack within 1 year), we evaluate the cumulative distribution function at x = 1:
CDF(1) = 1 - e^(-1/5) ≈ 0.1813
The probability that a randomly selected participant has experienced their first heart attack before the one-year follow-up visit is the complement of this probability:
P(first heart attack before 1 year) = 1 - CDF(1) ≈ 0.8187
Since the times to first heart attack for the participants are independent, the probability that at least three participants have experienced their first heart attack before the one-year follow-up visit can be calculated using the binomial distribution:
P(at least 3 participants with heart attack) = 1 - P(0 or 1 or 2 participants with heart attack)
Let's calculate this probability step-by-step:
P(0 participants with heart attack) = C(10, 0) * (0.8187)^0 * (1 - 0.8187)^(10 - 0) = 0.0249
P(1 participant with heart attack) = C(10, 1) * (0.8187)^1 * (1 - 0.8187)^(10 - 1) = 0.0833
P(2 participants with heart attack) = C(10, 2) * (0.8187)^2 * (1 - 0.8187)^(10 - 2) = 0.1630
P(at least 3 participants with heart attack) = 1 - (P(0 participants with heart attack) + P(1 participant with heart attack) + P(2 participants with heart attack))
P(at least 3 participants with heart attack) = 1 - (0.0249 + 0.0833 + 0.1630) ≈ 0.7297
Therefore, the probability that prior to the one-year follow-up visit, three or more of the participants have experienced their first heart attack is approximately 0.7297, or 72.97%.
To find the probability that three or more of the participants have experienced their first heart attack before the one-year follow-up visit, we can use the cumulative distribution function (CDF) of the exponential distribution.
The exponential distribution is defined by the parameter λ, which is equal to the reciprocal of the mean. In this case, the mean is 5, so λ = 1/5.
The CDF of the exponential distribution is given by the formula:
CDF(x) = 1 - e^(-λx)
Where x is the time (in this case, the time before the one-year follow-up), and e is the base of the natural logarithm (approximately 2.71828).
The probability that a specific participant has not experienced their first heart attack before the one-year follow-up is calculated as P(X > 1) where X is the time to first heart attack for that participant.
Since the times to first heart attack for the participants are independent, the probabilities for each participant are multiplied together.
Let's calculate the probability:
P(X > 1) = 1 - CDF(1) = 1 - (1 - e^(-λ * 1)) = e^(-λ * 1)
P(X > 1) = e^(-1/5) ≈ 0.8187
The probability that a specific participant has experienced their first heart attack before the one-year follow-up is 1 - P(X > 1) = 1 - 0.8187 ≈ 0.1813
Now, to calculate the probability that three or more participants have experienced their first heart attack before the one-year follow-up, we need to calculate the probability of having at most two participants experiencing their first heart attack.
P(at most 2 participants have their first heart attack) = P(0 participants) + P(1 participant) + P(2 participants)
P(at most 2 participants have their first heart attack) = (0.1813)^0 + (0.1813)^1 + (0.1813)^2
P(at most 2 participants have their first heart attack) ≈ 0.9823
Finally, to find the probability that three or more participants have experienced their first heart attack, we subtract this probability from 1.
P(three or more participants have their first heart attack) = 1 - 0.9823 ≈ 0.0177
Therefore, the probability that three or more of the participants have experienced their first heart attack since the initial assessment before the one-year follow-up is approximately 0.0177 or 1.77%.