Conditional probability question. Can someone check my work?

[wahaj]

Homework Statement

A new diagnostic test is developed in order to detect a particular disease. It is known that 1% of
the population has this disease. The diagnostic test is said to be 95% reliable. In other words, if
a person has this disease, the test will detect it 95% of the time. On the other hand, if a person
does not have the disease, the test will report that the person does not have the disease 90%
of the time.
A person is chosen at random from the population. The diagnostic test indicates that they have
the disease this test is designed to detect. What is the conditional probability that this person, in
fact, has the disease

Homework Equations

P(A|B)=P(A∩B)/P(B).

The Attempt at a Solution

See attached picture.
I am having trouble with conditional probability and I don't have a way of finding out whether the answer I am getting is correct. If I am doing this wrong can someone point out my mistake?

 

[Ray Vickson]

Homework Statement

A new diagnostic test is developed in order to detect a particular disease. It is known that 1% of
the population has this disease. The diagnostic test is said to be 95% reliable. In other words, if
a person has this disease, the test will detect it 95% of the time. On the other hand, if a person
does not have the disease, the test will report that the person does not have the disease 90%
of the time.
A person is chosen at random from the population. The diagnostic test indicates that they have
the disease this test is designed to detect. What is the conditional probability that this person, in
fact, has the disease

Homework Equations

P(A|B)=P(A∩B)/P(B).

The Attempt at a Solution

See attached picture.
I am having trouble with conditional probability and I don't have a way of finding out whether the answer I am getting is correct. If I am doing this wrong can someone point out my mistake?

Your answer is more than 10 times too large. I will not point out your errors, but will instead suggest an approach you can use if you find the formulas too difficult or confusing to use.

Imagine a population of N = 10,000,000 people. How many people have the disease? How many of the diseased people have a positive test result (that is, the test detects the disease)? How many of the non-diseased people give a positive test result? How many people altogether give a positive test result? Of all these positive-testers, how many actually have the disease? How can you get the required conditional probability from these last two items of data?

 

[haruspex]

You may have misunderstood the question. The conditional probability sought (I'm almost certain) is P[has the disease|test positive].
Your diagram is correct. From the diagram, what is P[has the disease & test positive]?

 

[wahaj]

Is this right?
How many people have the disease?-----10 000 000 * .01 = 100 000
How many of the diseased people have a positive test result?---- 100 000 * .95 = 95 000
How many of the non-diseased people give a positive test result?----- 9 900 000 * .10 = 990 000
How many people altogether give a positive test result?--------- 990 000+100 000 = 1 085 000
how many actually have the disease?------- 95 000/ 1 085 000 = 0.08755

 

[PeroK]

I'll leave you to do the maths, but here's how I'd look at it:

Only 1% of the population has the disease and the test is "pretty good" at getting a postive result in these cases. So, most of this 1% test positive.

But, 99% of the population do not have the disease and 10% of these will test positive. That's a lot.

Therefore, most of the people who test postive will NOT have the disease.

So: this is not a good test to use on the general population to screen for the disease. But, if you suspect a person has the disease, it's pretty good at confirming it.

 

[wahaj]

I'm getting an answer of 0.08755 which means that out of all the positive tests less than 10% has the disease. So this seems to sit well with your definition.

 

[Ray Vickson]

u

Is this right?
How many people have the disease?-----10 000 000 * .01 = 100 000
How many of the diseased people have a positive test result?---- 100 000 * .95 = 95 000
How many of the non-diseased people give a positive test result?----- 9 900 000 * .10 = 990 000
How many people altogether give a positive test result?--------- 990 000+100 000 = 1 085 000
how many actually have the disease?------- 95 000/ 1 085 000 = 0.08755

Your last line does not calculate what it says. The number who actually have the disease is already given as 100,000, from your first line. If you mean "what proportion of ... actually have the disease" then that is a different matter---and that is what you have computed in your last line. That, of course, is exactly what the words "conditional probability" mean in this case. So, aside from very bad working your result is OK now.

Do you see now how you could have used formulas right from the start, without going through the "population/sub-population" argument?

BTW: It would be better to write the last line using an approximation sign instead of an "=", because it is not true that 95000/1085000 = 0.08755 exactly, but it is true to 4 decimal places; so better to say ≈ 0.08755 instead. (To 20 digits we have ≈ 0.087557603686635944700, which rounds to 0.08756 to 5 decimal places).

 

[wahaj]

Yeah I got this. Thanks for the help