# Probability question

#1

Can anyone help with these questions?

For a)I) I did 1/1000 multiplied by 98/100 to get 98/100000 but I’m not convinced that this is correct?
0
3 months ago
#2
You need to use conditional probabiltiy as youre given that the test is positive.
0
#3
(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?
0
3 months ago
#4
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/prob...nditional.html
is a reasonable overview.
Last edited by mqb2766; 3 months ago
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#5
(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/prob...nditional.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?
0
3 months ago
#6
Again, conditional probability. Write it down clearly as per the previous link.
Its what you do when you create the simple tree.
Last edited by mqb2766; 3 months ago
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#7
(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.

Is this correct as a tree diagram?
0
3 months ago
#8
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?).
Last edited by mqb2766; 3 months ago
0
#9
(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?).
1
3 months ago
#10
Yes, hopefully you understand how the level two probabilities are really conditional probabilities so you can multiply along the branches to get the joint.
0
#11
(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?
0
3 months ago
#12
You need to use conditional probability, so
p(virus | test positive) = ...
0
#13
(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?
0
3 months ago
#14
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.
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#15
(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
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#16
(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?
0
3 months ago
#17
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?
0
3 months ago
#18
Mutually exclusive is different from independence. Two variables are independent if
p(A|B) = p(A)
So what would that mean for this problem?
0
#19
(Original post by mqb2766)
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?
I dont have textbooks, I just need someone to go throught a question like this with me
0
#20
(Original post by mqb2766)
Mutually exclusive is different from independence. Two variables are independent if
p(A|B) = p(A)
So what would that mean for this problem?
Im guessing that they are mutually exclusive?
0
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