# If a patient tests positive, what is the probability that they actually have the disease? [closed]

A hospital administers a test to see if a patient has a certain disease. Assume that we know the following three things:

5% of the overall population has the disease

If a person does have the disease, then the test has a 90% chance of correctly indicating that the person has it. (So 5% of the time, the test incorrectly indicates that the person doesn’t have the disease.)

If a person does not have the disease, then the test has a 15% chance of incorrectly indicating that the person has it; this is a “false positive” result. (So 90% of the time, the test correctly indicates that the person doesn’t have the disease.)

If a patient tests positive, what is the probability that they actually have the disease?

## closed as off-topic by Juho Kokkala, Michael Chernick, kjetil b halvorsen, Peter Flom♦Sep 30 '17 at 23:07

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Hugh has given you a good suggestion. This kind of question is a tipical Bayesian method in medicine.

Suppose $P(D)=5\%$ is the prevalence of the disease of the overall population.

We have the information $\text{the test has a 90% chance[probability] of correctly indicating that the person has it}$

We wrtite this as $P(+|D)=90\%$. This is also called $Senstivity$ in Epidemiology

We also have the informtion $\text{"If a person does not have the disease, then the test has a 15% chance of incorrectly indicating that the person has it"}$

$P(+|D^-)=15\%$

Now we have evrything to calcultate the probability $\text{"If a patient tests positive, what is the probability that they actually have the disease"}$

We write this sentence as $P(D|+)$

Now let us use Bayesian theorem to calculate this probability

$$P(D|+)=\frac{P(+|D)P(D)}{P(+)}=\frac{P(+|D)P(D)}{P(+| D)P(D)+P(+|D^-)P(D^-)}$$

Note for the denominator we use the law of total probability i.e $P(A)=P(A \cap B)+P(A \cap B^c)$

You have evrything to calculate the probability.

$P(+|D)=90\% \\ P(D)=5\% \\ P(+|D^-)=15\%\\P(D^-)=1-5\%=95\%$

$$\therefore P(D|+)=\frac{0.9*0.05}{0.9*0.05+0.15*0.95}=0.24$$

This means even you are test postive your chance to have the disease is $24 \%$ which is quite low.

Suppose 200 people take the test. Fill in the expected numbers in this table:  | diseased | healthy | total tests positive | ? | ? | ? tests negative | ? | ? | ? total | ? | ? | 200  The answer is then  # (tests positive and diseased) / total # tests positive 

• Thank you so much I figured, was a bit confused because of the wrong percentages between the bracket but got it now. Thanks a lot again – user179033 Sep 30 '17 at 22:06