# Chance of getting sick

1. Feb 19, 2006

### haynewp

Say I were exposed to someone who has the flu. And the chance of me getting sick were 50% from being exposed to that person.

Now, say 50% of the people who actually have the flu virus develop symptoms within the first 2 days. I make it up to day 3 without any symptoms. I think I should be able to cut my odds again that I did not contract the flu from this person: 1/2*1/2=1/4 chance I may have the flu?

2. Feb 19, 2006

### 0rthodontist

Okay, I got my answer: you can't cut the odds to 1/4. Once you know you have a 1/2 chance of having the flu, you can be viewed as a person selected at random from the following population:
A B
C D
Where C and D have the flu.
Now let's say you know that only D shows symptoms. Now what is your chance of having the flu, given that you do not show symptoms?

Last edited: Feb 19, 2006
3. Feb 20, 2006

### haynewp

Alright.

If I word it to where 99.999% of people show symptoms within the first 2 days, and I make it to day 3 without any synptoms, it shouldn't make any difference. I would still be at the original 50% chance.

But it sure would seem like I would be in a lot better shape than the original 1 in 2 odds, since I had made it to day 3. I guess you're right.

4. Feb 20, 2006

### qbert

I was playing around with this and it looks like you
need one more piece of information.

So let me change the problem slightly
Let's say you have a 50% chance of getting
sick. And if you're sick you have a 50% chance
of showing symptoms in under two days.
and if you don't get sick you never show
the symptoms in under two days (no false positives).

Then: Let A be the event you get sick
and B be the event you show symptoms in under two days.

Then you are given P(A)=.5, P(B|A)=.5, and P(B|A^c)=0
And you want to find out what is P(A|B^c) {you are
sick but didn't show symptoms in under two days}

1. P(A|B^c)P(B^c) = P(B^c|A)P(A)
2. P(B^c|A) = 1 - P(B|A)
3. P(B^c) = 1-P(B) = 1- P(B|A^c)P(A^c) - P(B|A)P(A)

thus P(A|B^c) = (1-P(B|A))P(A) / (1 - P(B|A^c)P(A^c) - P(B|A)P(A) )

or putting in numbers:
(1- .5)(.5) /( 1 - 0 - (.5)(.5)) = .5^2/(1-.5^2) = 1/3

So, if you make it to day three you have a 1 in 3 chance of being
sick.

5. Feb 20, 2006

### 0rthodontist

This conclusion was already clear from my A B C D population.

Last edited: Feb 20, 2006
6. Feb 20, 2006

### haynewp

"Okay, I got my answer: you can't cut the odds to 1/4. Once you know you have a 1/2 chance of having the flu, you can be viewed as a person selected at random from the following population:
A B
C D
Where C and D have the flu.
Now let's say you know that only D shows symptoms. Now what is your chance of having the flu, given that you do not show symptoms?"

If A is defined as not being sick, and C is sick with no symptoms, then what is B? It's been a while since I did probabilities. It must be that C and D are subsets of B?

7. Feb 20, 2006

### qbert

The only point I was making is that
the problem is incompletely specified.

To actually solve it you need to make additional
assumptions. explicitly, you need to know
the probablility of showing symptoms while
not being sick. (Or anything, from which you
can infer it.)

For Example. If in one extreme,
showing the symptoms is independent
of being sick. then you're chance of
being sick is 1/2 whether or not you show
symptoms.

In the other extreme you show no
symptoms while being sick. Then the
probabillity of being sick after 2 days varies
from 0 to 1/3 depending on what percenctage
of sickies show symptoms inside of 2 days.

8. Feb 20, 2006

### qbert

Take the setup. Probability of being sick = 1/2
If sick, the probabillity of showing symptoms within 2 days = 2/3
If not sick, the prob. of showing symptoms within 2 days = 0

then by the same analysis that i gave
the probabillity of being sick after two days of not showing symptoms
is 1/4.

9. Feb 20, 2006

### 0rthodontist

We are given that the probability of showing symptoms within 2 days if you are sick is 1/2... yes, you are correct, if you change the problem entirely you can reduce the total chance of being sick to 1/4 or whatever fraction you want less than 1/2, but usually we try to solve problems without arbitrarily altering the values given.

It can easily be inferred that if no sick people show symptoms within 2 days then your probability of being sick after 2 days of no symptoms is 1/2.

I assume you mean in the first sentence of the quote, "In the other extreme you show no symptoms without being sick."

It seems you may not have understood my A, B, C, D population. If you do not show symptoms then you are not D, therefore you are A, B, or C. Only C has the flu therefore your chance of having the flu is 1/3.

Last edited: Feb 20, 2006
10. Feb 20, 2006

### qbert

I was making two points.

1st, To get the answer you have to assume an extra piece of
information that isn't stated anywhere. (And may not even
be a good assumption.)

2nd, I was trying to clear the confusion, which I perceived in post
3, by giving a general derivation. Followed with a couple of
different examples (ie what is the setup to reduce the chance to 1/4).

.... as to the typo. C'est La Vie. Thanks for the correction.

11. Feb 20, 2006

### haynewp

I see, A and B are there just to represent the 50% that will not get sick in the original population A,B,C,D. (There is no difference between A and B.)

You are just left with A (well), B (well) or C (sick-no symptoms) if you do not show symptoms (D).