Probability problem confirmation needed, help!

[johammbass]

Hi all,

The problem is as follows:

A certain disease can be detected by a blood test in 95% of those who have it. Unfortunately, the test also has a 0.02 probability of showing that a person has the disease when in fact he or she does not. It has been estimated that 1% of those people who are routinely tested actually have the disease. If the test shows that a certain person has the disease, find the probability that the person actually has it.

This is how I went about it:
For straight forwardness we will look at the population as being 10000 people.

95% of those who have it will test "yes" i.e. 95% of 1% = 0.95 x 0.01 = 0.0095 from all population, so from 10000 people it would be 95 people

2% of those who do not have it will also test "yes" i.e. 2% of 99% = 0.02 x 0.99 = 0.0198 from all population, so from 10000 people it would be 198 people

So, the total number of all "yes" tests (positive "yes" and negative "yes") would be 95+198=293 "yes" tests

Since we know that 1% of population actually have the disease i.e. 100 people then we can conclude that chance of having disease if the test was positive is 100/293=0.3413


My answer came out as 0.3413. But the lecturer's answer came out as 0.324
Which one of us is right? Maybe both are wrong?


Thank you in advance!

 

[wjm11]

A certain disease can be detected by a blood test in 95% of those who have it. Unfortunately, the test also has a 0.02 probability of showing that a person has the disease when in fact he or she does not. It has been estimated that 1% of those people who are routinely tested actually have the disease. If the test shows that a certain person has the disease, find the probability that the person actually has it.

This is how I went about it:
For straight forwardness we will look at the population as being 10000 people.

95% of those who have it will test "yes" i.e. 95% of 1% = 0.95 x 0.01 = 0.0095 from all population, so from 10000 people it would be 95 people

2% of those who do not have it will also test "yes" i.e. 2% of 99% = 0.02 x 0.99 = 0.0198 from all population, so from 10000 people it would be 198 people

So, the total number of all "yes" tests (positive "yes" and negative "yes") would be 95+198=293 "yes" tests

Since we know that 1% of population actually have the disease i.e. 100 people then we can conclude that chance of having disease if the test was positive is 100/293=0.3413


My answer came out as 0.3413. But the lecturer's answer came out as 0.324

The lecturer's answer is correct. (You were doing fine until your last step.) I recommend a "Probability Tree Diagram" for a problem like this. You might research that. Anyway, the solution is

P(those who have the disease and test positive)/P(those who test positive) = [(.01)(.95)]/[((.01)(.95) + (.99)(.02)] = .3242

 

[soroban]

Hello, johammbass!

A certain disease can be detected by a blood test in 95% of those who have it.
Unfortunately, the test also has a 2% probability of showing that a person has the disease when in fact he or she does not.
It has been estimated that 1% of those people who are routinely tested actually have the disease.
If the test shows that a certain person has the disease, find the probability that the person actually has it.

\(\displaystyle \text{We want: }\(\text{has disease}\,|\,\text{tests positive})\)

\(\displaystyle \text{Bayes' Theorem: }\;P(\text{has disease}\,|\text{tests positive}) \;=\;\frac{P(\text{has disease}\,\wedge\,\text{tests positive})}{P(\text{tests positive})}\)


\(\displaystyle \text{The denominator has two cases:}\)

. . \(\displaystyle \text{Case 1: }\(\text{has disease}\,\wedge\,\text{tests positive}) \;=\;(0.01)(0.95) \:=\:0.0095\)

. . \(\displaystyle \text{Case 2: }\(\text{not disease}\,\wedge\,\text{tests positive}) \;=\;(0.99)(0.02) \:=\:0.0198\,\)

\(\displaystyle \text{Hence, the denominator is: }\:0.0095 + 0.0198 \;=\;0.0293\)


\(\displaystyle \text{The numerator is case [1]: }\(\text{has disease}\,\wedge\text{tests positive}) \:=\:0.0095\)


\(\displaystyle \text{Therefore: }\(\text{has disease}\,|\,\text{tests positive}) \;=\;\frac{0.0095}{0.0293} \;=\;0.324232082 \;\approx\;0.324\)