Probability question

You need to use conditional probabiltiy as youre given that the test is positive.

Reply 2

OP

Original post by mqb2766

You need to use conditional probabiltiy as youre given that the test is positive.

Okay, so given that 1/1000 is positive 98/100 actually have the virus?

Reply 3

sort of, you're told they have tested positive, so that can come from a virus,correct test
or
not virus, false test

Why not draw the simple tree and write down the equation you need to use and keep going until you get stuck?
https://www.mathsisfun.com/data/probability-events-conditional.html
is a reasonable overview.

(edited 2 years ago)

Reply 4

OP

Original post by mqb2766

sort of, you're told they have tested positive, so that can come from a virus,correct test
or
not virus, false test

Why not draw the simple tree and write down the equation you need to use and keep going until you get stuck?
https://www.mathsisfun.com/data/probability-events-conditional.html
is a reasonable overview.

I understand that I need to use 1% is false positive but I’m not sure how to because our 1/1000 people test positive and 98/100 actually have the virus, why can I just multiply them to get the percent of people that test positive and have virus?

Reply 5

Again, conditional probability. Write it down clearly as per the previous link.
Its what you do when you create the simple tree.

(edited 2 years ago)

Reply 6

OP

Original post by mqb2766

Again, conditional probability. Write it down clearly as per the previous link.
Its what you do when you create the simple tree.

FF43F5AB-C039-4CFA-AFA8-B62764ECD51B.jpg.jpeg

Is this correct as a tree diagram?

Reply 7

No. If you have the virus, its impossible to get a false positive.

The term Positive/Negative at the first level don't distinguish the virus or the test. It must be
* first level is the virus or not
* second level is test positive or negative

If you're getting it incorrect at this level, it really would be worth going through some worked examples carefully (textbook?).

(edited 2 years ago)

Reply 8

OP

Original post by mqb2766

No. If you have the virus, its impossible to get a false positive.

The term Positive/Negative at the first level don't distinguish the virus or the test. It must be
* first level is the virus or not
* second level is test positive or negative

If you're getting it incorrect at this level, it really would be worth going through some worked examples carefully (textbook?).

B1D0B5FA-D72B-42C3-A87D-264424CD3969.jpg.jpeg

Reply 9

Yes, hopefully you understand how the level two probabilities are really conditional probabilities so you can multiply along the branches to get the joint.

Reply 10

OP

Original post by mqb2766

Yes, hopefully you understand how the level two probabilities are really conditional probabilities so you can multiply along the branches to get the joint.

So if a person tests positive for the virus and has the virus, that is 98/100 and then 1/1000 people have the virus, so do I just not multiply it since the the probability of someone having the virus is 0.001 and then testing positive is 0.98?

Reply 11

You need to use conditional probability, so
p(virus | test positive) = ...

Reply 12

OP

Original post by mqb2766

You need to use conditional probability, so
p(virus | test positive) = ...

P(test positive and virus) / P(test positive) but aren't they independent?

Reply 13

Id really suggest trying to work through issues like these yourself first. Why do you think they're independent, how can you determine, ... Tbh, you should be able to just spot this from the tree and the knowledge isnt really beyond gcse. Though there are a few ways to resolve the issue.

Reply 14

OP

Original post by mqb2766

Id really suggest trying to work through issues like these yourself first. Why do you think they're independent, how can you determine, ... Tbh, you should be able to just spot this from the tree and the knowledge isnt really beyond gcse. Though there are a few ways to resolve the issue.

okay i give up, sorry for wasting your time

Reply 15

OP

Original post by mqb2766

Id really suggest trying to work through issues like these yourself first. Why do you think they're independent, how can you determine, ... Tbh, you should be able to just spot this from the tree and the knowledge isnt really beyond gcse. Though there are a few ways to resolve the issue.

if P(a|b) = P(AnB)/P(A) is this just for mutually exclusive, because if it was independent then it just cancels for P(B) right?

Reply 16

You're not really using knowlege of conditional probabilities which what the question relies on.
Do you have the textbook, as there willl be some worked examples in there?

Reply 17

Mutually exclusive is different from independence. Two variables are independent if
p(A|B) = p(A)
So what would that mean for this problem?

Reply 18

OP