Conditional probability of a test records this positive result

[songoku]

Homework Statement

Please see below

Relevant Equations

$P(A|B)=\frac{P(A \cap B)}{P(B)}$

My attempt:
$$P(\text{B is positive}|\text{A is positive})=\frac{P(\text{B is positive} \cap \text{A is positive})}{P(\text{A is positive})}$$
$$=\frac{P(\text{B is positive})\times P(\text{A is positive})}{P(\text{A is positive})}$$
$$=P(\text{B is positive})$$
$$=0.01 \times 0.99 + 0.99 \times 0.02$$
$$=0.0297$$

Where is my mistake?

Thanks

 

[Office_Shredder]

You assumed that A and B being positive are independent events, which is not true. If A is positive, B is much more likely to be positive. I suggest starting off by computing the probability the person has the disease if A come back positive.

 

[songoku]

You assumed that A and B being positive are independent events, which is not true. If A is positive, B is much more likely to be positive. I suggest starting off by computing the probability the person has the disease if A come back positive.

$$P(\text{has disease}|\text{A is positive})=\frac{P(\text{A is positive}\cap \text{has disease})}{P(\text{A is positive})}$$
$$=\frac{0.01 \times 0.99}{0.01 \times 0.99 + 0.99 \times 0.04}$$
$$=\frac{97}{493}$$

What would be the next step? Thanks

 

[Office_Shredder]

Now compute P(B is positive ) given that is the probability the person has the disease.

 

[songoku]

Now compute P(B is positive ) given that is the probability the person has the disease.

Do you mean this:
$$P(\text{B is positive}|\text{has disease})=0.99$$

or

$$P(\text{has disease}|\text{B is positive})=\frac{0.01 \times 0.99}{0.01 \times 0.99 + 0.99 \times 0.02}$$
$$=\frac 1 3$$

 

[Office_Shredder]

After you get the A positive test, you know they are 97/493 to be positive. Given that, what are the odds the B test is positive?

You have to combine the odds that B is positive given they have the disease, and that B is positive given they don't have the disease.

 

[songoku]

After you get the A positive test, you know they are 97/493 to be positive. Given that, what are the odds the B test is positive?

You have to combine the odds that B is positive given they have the disease, and that B is positive given they don't have the disease.

$$P(\text{B is positive}|\text{has disease})=0.99$$

$$P(\text{B is positive}|\text{does not have disease})=0.02$$

But sorry I don't how to relate those to find the answer.

Do I also need to consider P(does not have disease | A is positive)?

Thanks

 

[Office_Shredder]

You just want to use P(B is positive) = P(B is positive | has disease) P(has disease) + P(B is positive | doesn't have disease) P(doesn't have disease)

 

[songoku]

You just want to use P(B is positive) = P(B is positive | has disease) P(has disease) + P(B is positive | doesn't have disease) P(doesn't have disease)

Wouldn't it be the same if I just calculate P(B is positive) without using conditional probability?

P(B is positive) = P(has disease and B is positive) + P(does not have disease and B is positive)
= 0.01 x 0.99 + 0.99 x 0.02
= 0.0297

Sorry but I still can't relate all the hints to find P(A is positive and B is positive).

Thanks

 

[Office_Shredder]

Wouldn't it be the same if I just calculate P(B is positive) without using conditional probability?

It's different because I'm telling you to use a different probability that the person has the disease than the starting one. Use the information you learned from getting the A test being positive.

This person does not have a 1% chance is having the disease.

0.01 x 0.99

So you should not have this showing up anywhere.

 

[songoku]

It's different because I'm telling you to use a different probability that the person has the disease than the starting one. Use the information you learned from getting the A test being positive.

This person does not have a 1% chance is having the disease.
So you should not have this showing up anywhere.

P(B has disease) = 0.99 x 97/493 + 0.02 x (1 - 97/493) = 21.1%

Is this what you mean?

Thanks

 

[haruspex]

P(B has disease) = 0.99 x 97/493 + 0.02 x (1 - 97/493) = 21.1%

Is this what you mean?

Thanks

That's the answer I get, but I don’t follow how you got it.
If A is the event test A is +ve, B is the event test B is +ve, and D is the event the patient has the disease then I think of it as
P(A)=P(D)P(A|D)+P(-D)P(A|-D)
P(A&B)=P(D&A&B)+P(-D&A&B)
P(B|A)=P(A&B)/P(A)

 

[songoku]

That's the answer I get, but I don’t follow how you got it.
If A is the event test A is +ve, B is the event test B is +ve, and D is the event the patient has the disease then I think of it as
P(A)=P(D)P(A|D)+P(-D)P(A|-D)
P(A&B)=P(D&A&B)+P(-D&A&B)
P(B|A)=P(A&B)/P(A)

The approach suggested by @Office_Shredder is to find P(D | A) and use this as the probability of getting the disease.

So when finding P(B|A) it becomes something like P(B ∩ (D|A)) + P(B ∩ (D|A)')

 

[Office_Shredder]

I think at some level trying to write down generic formulas becomes less helpful. The question asks you to find the probability that B is positive, which you can only figure out if you know whether someone has the disease, so first you should figure out if they have the disease or not, and then use that to figure out if B is positive.

Also, (D|A)' probably should be (D'|A).

 

[haruspex]

The approach suggested by @Office_Shredder is to find P(D | A) and use this as the probability of getting the disease.

So when finding P(B|A) it becomes something like P(B ∩ (D|A)) + P(B ∩ (D|A)')

Not sure how to read P(B ∩ (D|A)). D|A is not an event. Do you mean P((B ∩ D)|A)) + P((B ∩ D)|A'))? But that just reduces to P(B ∩ D). I guess you mean P(B|A)= P((B ∩ D)|A)) + P((B ∩ D')|A))?

Anyway, finding P(D|A) seems like a long way round to me. I'd need to see the whole solution to compare.

 

[songoku]

Not sure how to read P(B ∩ (D|A)). D|A is not an event. Do you mean P((B ∩ D)|A)) + P((B ∩ D)|A'))? But that just reduces to P(B ∩ D). I guess you mean P(B|A)= P((B ∩ D)|A)) + P((B ∩ D')|A))?

Yes, I think that's the one, P(B|A)= P((B ∩ D)|A)) + P((B ∩ D')|A))

Anyway, finding P(D|A) seems like a long way round to me. I'd need to see the whole solution to compare.

$$P(D|A)=\frac{P(D \cap A)}{P(A)}$$
$$=\frac{0.01 \times 0.99}{0.01 \times 0.99 + 0.99 \times 0.04}$$
$$=\frac{97}{493}$$

P(B|A) = P((B ∩ D)|A)) + P((B ∩ D')|A)) = 0.99 x 97/493 + 0.02 x (1 - 97/493) = 21.1%

I guess the working is something like that.

Thanks

 

[haruspex]

Yes, I think that's the one, P(B|A)= P((B ∩ D)|A)) + P((B ∩ D')|A))
$$P(D|A)=\frac{P(D \cap A)}{P(A)}$$
$$=\frac{0.01 \times 0.99}{0.01 \times 0.99 + 0.99 \times 0.04}$$
$$=\frac{97}{493}$$

P(B|A) = P((B ∩ D)|A)) + P((B ∩ D')|A)) = 0.99 x 97/493 + 0.02 x (1 - 97/493) = 21.1%

I guess the working is something like that.

Thanks

Ok, about the same length as my approach in post #12.

 

[songoku]

Thank you very much Office_Shredder and haruspex