Question 1182617: . Consider the following data for Alzheimer disease prevelance in a certain community among elderly.
> Alzheimer=data.frame(
+ AgeClass=c("65-69","70-74","75-79","80-84","85+"),
+ Males=c(1.6,0.0,4.9,8.6,35.0),
+ Females=c(0.0,2.2,2.3,7.8,27.9),
+ row.names="AgeClass"
+ )
Find the probability that among two randomly selected women one having age between 70 and 74 while the other more than 85 and three men with age between 65 and 69 none will have Azheimer disease. We assume that individuals are selected independently from each age category as indicated.
2. What is the probability that at least one of the women has the disease?
3. What is the probability that at least one of the men has the disease?
What is the probability that at least one of the five has the disease?
5. What is the probability that exactly one person out the five has the disease?
6. If we know that exactly one person has the disease, what is the probability that this is not a woman?
7. If we know that three persons have the disease, what is the probability that all are men?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to calculate the probabilities, explained step by step:
First, let's calculate the probabilities of *not* having Alzheimer's for each group:
* **Women 70-74:** 1 - 0.022 = 0.978
* **Women 85+:** 1 - 0.279 = 0.721
* **Men 65-69:** 1 - 0.016 = 0.984
**1. Probability that both women are disease-free and all three men are disease-free:**
Since the selections are independent, we multiply the probabilities:
(0.978) * (0.721) * (0.984) * (0.984) * (0.984) ≈ 0.677
**2. Probability that at least one woman has the disease:**
This is the complement of *neither* woman having the disease:
1 - (0.978 * 0.721) ≈ 0.298
**3. Probability that at least one man has the disease:**
This is the complement of *none* of the men having the disease:
1 - (0.984 * 0.984 * 0.984) ≈ 0.048
**4. Probability that at least one of the five has the disease:**
This is the complement of *none* of the five having the disease:
1 - (0.978 * 0.721 * 0.984 * 0.984 * 0.984) ≈ 0.323 (which is 1-0.677)
**5. Probability that exactly one person out of the five has the disease:**
We need to consider each case (one of the five has the disease, the others are healthy) and sum the probabilities:
* Woman 70-74: (0.022 * 0.721 * 0.984 * 0.984 * 0.984) ≈ 0.020
* Woman 85+: (0.978 * 0.279 * 0.984 * 0.984 * 0.984) ≈ 0.264
* Man 65-69 (1st): (0.978 * 0.721 * 0.016 * 0.984 * 0.984) ≈ 0.001
* Man 65-69 (2nd): (0.978 * 0.721 * 0.984 * 0.016 * 0.984) ≈ 0.001
* Man 65-69 (3rd): (0.978 * 0.721 * 0.984 * 0.984 * 0.016) ≈ 0.001
Summing these probabilities: 0.020 + 0.264 + 0.001 + 0.001 + 0.001 ≈ 0.287
**6. Probability that it's not a woman, given exactly one person has the disease:**
We need the conditional probability: P(Not a woman | Exactly one has the disease). This is equal to the probability that exactly one man has the disease divided by the probability that exactly one person has the disease.
(0.001 + 0.001 + 0.001) / 0.287 ≈ 0.003/0.287 ≈ 0.010
**7. Probability that all three are men, given three people have the disease:**
We need the conditional probability: P(All three are men | Three have the disease). We need to calculate the probability that three men have the disease and divide it by the probability that three people have the disease. This is tricky because we are not given enough information to calculate the probability that three people have the disease. We only know the probability that a specific individual has the disease, not the probability that a group of three have the disease. We need more information on the dependency structure between different individuals. Assuming they are independent, the probability that the three men all have the disease is (0.016)^3. We also need the probabilities of various combinations of 3 people having the disease (e.g. 2 men + 1 woman, etc). We can't calculate that from the data provided. So we can't calculate the conditional probability.
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