Twenty-five percent of all deaths (all ages) are caused by diseases of the heart. Ischemic heart disease accounts for 16.4% of all deaths and heart failure for 2.3%. Choose one death at random. What is the probability that it is from ischemic heart disease given that it was from heart disease? choose two deaths at random: what is the probability that at least one is from heart disease?
To find the probability that a death is from ischemic heart disease given that it is from heart disease, we need to use conditional probability.
Let's represent the events:
A: Death is from ischemic heart disease
B: Death is from heart disease
We are given that P(B) = 25% (since 25% of all deaths are due to heart disease), and P(A|B) = 16.4% (since 16.4% of all deaths are due to ischemic heart disease).
The probability that a randomly chosen death is from ischemic heart disease, given that it is from heart disease, is given by P(A|B) = P(A ∩ B) / P(B). Using this formula, we can calculate:
P(A ∩ B) = P(A|B) * P(B) = 0.164 * 0.25 = 0.041 (or 4.1%)
Therefore, the probability that a randomly chosen death is from ischemic heart disease, given that it is from heart disease, is 4.1%.
Now, to find the probability that at least one of two deaths is from heart disease, we can use the complement rule.
Let's represent the events:
C: At least one death is from heart disease
We want to find P(C'). The probability that none of the two deaths are from heart disease is given by P(C') = 1 - P(C).
Using the complement rule:
P(C') = 1 - P(C)
To find P(C), we need to find the probability that both deaths are not from heart disease. Since the probability of a death being from heart disease is 25%, the probability of it not being from heart disease is 75% (or 0.75).
P(C) = P(both deaths not from heart disease) = 0.75 * 0.75 = 0.5625 (or 56.25%)
Therefore, P(C') = 1 - P(C) = 1 - 0.5625 = 0.4375 (or 43.75%)
So, the probability that at least one of two deaths is from heart disease is 43.75%.
To find the probability that a death is from ischemic heart disease given that it was from heart disease, we need to use conditional probability.
Conditional probability is calculated using the formula:
P(A|B) = P(A ∩ B) / P(B)
Where P(A|B) is the probability of event A occurring given that event B has occurred, P(A ∩ B) is the probability of both events A and B occurring, and P(B) is the probability of event B occurring.
In this case, event A is a death from ischemic heart disease, and event B is a death from heart disease.
Given that 25% of all deaths are caused by heart disease, the probability of event B occurring is 0.25.
Given that 16.4% of all deaths are caused by ischemic heart disease, the probability of event A occurring is 0.164.
To find the probability of both events A and B occurring (P(A ∩ B)), we can multiply the probability of event B by the probability of event A given B:
P(A ∩ B) = P(A|B) * P(B) = 0.164 * 0.25 = 0.041
Now, we can substitute these values into the conditional probability formula to find the probability that a death is from ischemic heart disease given that it was from heart disease:
P(A|B) = P(A ∩ B) / P(B) = 0.041 / 0.25 = 0.164
Therefore, the probability that a death is from ischemic heart disease given that it was from heart disease is 0.164, or 16.4%.
Moving on to the second question, to find the probability that at least one death is from heart disease when choosing two deaths at random, we need to find the complement of the event that neither death is from heart disease.
The probability that neither death is from heart disease can be calculated as follows:
P(neither death is from heart disease) = (1 - 0.25)^2 = 0.5625
Therefore, the probability that at least one death is from heart disease can be found by subtracting the probability of the complement event from 1:
P(at least one death is from heart disease) = 1 - 0.5625 = 0.4375
Thus, the probability that at least one death is from heart disease is 0.4375, or 43.75%.