Simple problems

[Peytonator]

Hi there,

I am having a problem with the following with the following two problems:

1. "In a survey conducted on the men above 50 years of age in a certain community it was determined that 1 in 4 of this group smoked, that 3 in 4 of those who had lung cancer were also smokers, and that 1 in 300 had lung cancer. Determine (a) the probability that a smoker of this age group has lung cancer and (b) that probability that a man above 50 is either a smoker or has lung cancer."

2. "Tests employed in the detection of a paticular disease are 90% effective; they fail to detect it in 10% of cases. In persons free of the disease, the tests indicate 1% to be affected and 99% to not be affected. From a large population, in which only 0.2% have the disease, one person is selected at random, is given the test, and presence of the disease is indicated. What is the probability that the person is affected."

I've tried answering them both but get dismally wrong answers. Please help!

Correct answers:
1. 1/100; 301/1200
2. 90/589

Many thanks!

* EDIT: sorry I put this topic in the wrong section!

[AtlasFrysmith]

The first step in dealing with any probability problem is to define your events. You'll see why in a bit. Let's just deal with the first question.
Let
S=(The person selected is a smoker)
We are told that P(S)=1/4

C=(The person selected has cancer)
We are told that P(C)=1/300


We also have an event "the person selected is a smoker GIVEN THAT they have cancer." It's in slightly different words in the problem, but that's what it's saying. Why did I reword it? So that we can easily write the event as
(S|C). This is read "S given C"
And we're told that P(S|C)=3/4.

Now, what are we trying to find? In part a), it's P(C|S). Think about why this is if it's not clear. In part b), it's P(S U C). Again, make sure you understand why this is the probability we're interested in. If you don't, let me know and we can talk about that.

OK, so why all the letters instead of words? Well, your formulas in your book are all in letters, so now we just need to find relationships between the things we know, P(S), P(C), and P(S|C), and the things we don't know, P(C|S) and P(S U C).

I'll leave this part to you--look carefully through your book and you should be able to find a formula for each one. The reason I want you to do it this way is because it's good practice to turn everything into symbolic language--you'll realize that most of these problems are basically the same in structure; only the numbers change.

[talkstatsdkf]

One suggestion is to pick a number N and convert all the probabilities to numbers. It can be easier then to visualize exactly what the proportions are for each designation.

For example, a problem might say that 1/4 of seniors take Calculus, 1/5 of seniors take Statistics, and 1/10 of seniors take both Calculus and Statistics.

Assume there are 100 seniors:

25 take Calculus
20 take Statistics
10 take both Calculus and Statistics

the problem could then ask how many take Calculus or Statistics:

25 + 20 - 10 = 35 students

or, equivalenty:

1/4 + 1/5 - 1/10 = 7/20

100*7/20 = 35

David

[Peytonator]

Thanks for these replies I get correct answers with:

a) P(S & C) = P(S).P(C|S) = P(C).P(S|C)

b) P(S or C) = P(S) + P(C) - P(S).P(C|S)

But now in (2) if I say

P(D) = P(disease)
P(E) = P(effective)
P(F) = P(non-effective)

P(D & E) = P(D).P(E|D) = (0.02)(0.9) = 9/500 which is wrong

How do I do this?

[AtlasFrysmith]

You need to define your events differently, and you're solving for the wrong thing.

Let X=(Random patient tests positive)
and Y=(Random patient has disease)

What are you trying to find? (It's a conditional probability, not a probability of an intersection)
What do you know from the problem? (Hint: you know lots of other conditional probabilities and P(Y)).

Write all the probabilities that you're given, and then see if you can manipulate them into the one you need. You'll need to use all the information given.

I'm not trying to be a jerk by leading you through this step by step, but I think it'll help you learn the method better than just working the whole example at once. And trust me, it's a good habit to write out all the events and the probabilities at the beginning.

[Peytonator]

It's a conditional probability, not a probability of an intersection

What do you mean by this?

I'm still pretty much lost. All I can get to is:

P(D) = P(disease) = 0.02
P(E) = P(effective) = 0.9
P(NE) = P(non-effective) = 0.1
P(H-E) = P(effective test in a healthy person) = 0.99
P(H-NE) = P(non-effective test in a healthy person) = 0.01

I don't seem to see the links yet. How does this come in:

Let X=(Random patient tests positive)
and Y=(Random patient has disease)

[AtlasFrysmith]

What you're trying to find is the probability that a person who tests positive actually has the disease. Call this P(Y|X)--probability of disease given positive test. You know P(Y)--probability of disease, you DON'T yet know P(X)--probability of positive test, but you do know P(X|Y), P(X|Yc), P(Xc|Y), and P(Xc|Yc).

Xc is the complement of X, so the event that the test is negative, and Yc is the complement of Y, or the event that the person does not have the disease.

Notice how I do these in terms of whether the test is positive, not whether the test is right. That's where you got tripped up, I think.

Given what you know, you should be able to manipulate things to get what you need.

Here's a hint--remember that P(X)=P(X|Y)P(Y) + p(X|Yc)P(Yc).