Conditional Probability / Baye's Thm

My text is lacking a clear example on this one.

Given the following information

Code:

` Prevalence of Alzheimer’s disease (cases per 100)`

Age Group Males Females

65-69 1.6 0

70-74 0 2.2

75-79 4.9 2.3

80-84 8.6 7.8

85+ 35 27.9

Suppose an unrelated 77 year old man, a 76 year old woman and an 82 year old woman are selected from a community.

**1) What is the probability that all three have Alzheimer’s disease?**

Let A = 77 year old man has Alzheimer’s

Let B = 76 year old woman has Alzheimer’s

Let C = 82 year old woman has Alzheimer’s

I say the answer is P(A) x P(B) x P(C) = 4.9/100 x 2.3/100 x 7.8/100

**2) What is the probability that at least one of the women have Alzheimer’s disease?**

Do I need to apply Baye's thm here? Or is it some binomial thing where I find p = 2.3/100 x 7.8/100

**3) What is the probability that at least one of the three have Alzheimer’s disease?**

If my method for Q2 is correct I would try something similar here.

Thanks for your interest...(Happy)

Re: Conditional Probability / Baye's Thm

Quote:

Originally Posted by

**Bushy** My text is lacking a clear example on this one.

Given the following information Code:

` Prevalence of Alzheimer’s disease (cases per 100)`

Age Group Males Females

65-69 1.6 0

70-74 0 2.2

75-79 4.9 2.3

80-84 8.6 7.8

85+ 35 27.9

Suppose an unrelated 77 year old man, a 76 year old woman and an 82 year old woman are selected from a community.

**1) What is the probability that all three have Alzheimer’s disease?**
Let A = 77 year old man has Alzheimer’s

Let B = 76 year old woman has Alzheimer’s

Let C = 82 year old woman has Alzheimer’s

I say the answer is P(A) x P(B) x P(C) = 4.9/100 x 2.3/100 x 7.8/100

**2) What is the probability that at least one of the women have Alzheimer’s disease?**
Do I need to apply Baye's thm here? Or is it some binomial thing where I find p = 2.3/100 x 7.8/100

**3) What is the probability that at least one of the three have Alzheimer’s disease?**
If my method for Q2 is correct I would try something similar here.

Thanks for your interest...(Happy)

The three are (assumed) independent. Bayes only enters in the calculation of them individually having Alzteimer's, which are the individual cell probabilities from the table (as you have used). So your answer to 1 is correct.

The probability that at least one of the women has A. is one minus the probability that neither have it so is 1- [(1-2.3/100) x (1-7.8/100)]

The third part now uses the method I used above for the two women, first find the probability that none have A. then subtract that from 1.

CB