Hey everyone..this is my firts post..I can't solve this problem..can somebody help me pls..

1.) Assume that a test to detect a disease whose prevalence is (1/1000) has a false positive rate of 5% and a true positive rate of 100%. What is the probability that a person found to have a positive result actually has the disease, assuming that you know nothing about the person's symptoms?

Define events: D= the event that the person has the disease
T= the event that the test result is positive.
Given:

P(D)=0.001
P(T|D complement)=0.05
P(T|D)=1.00

please can anybody help..i'll really appreciate it..

2. Hello,
Originally Posted by takeyourmark
Hey everyone..this is my firts post..I can't solve this problem..can somebody help me pls..

1.) Assume that a test to detect a disease whose prevalence is (1/1000) has a false positive rate of 5% and a true positive rate of 100%. What is the probability that a person found to have a positive result actually has the disease, assuming that you know nothing about the person's symptoms?

Define events: D= the event that the person has the disease
T= the event that the test result is positive.
Given:

P(D)=0.001
P(T|D complement)=0.05
P(T|D)=1.00

please can anybody help..i'll really appreciate it..
So you're looking for P(D|T)

$\mathbb{P}(D|T)=\frac{\mathbb{P}(D \cap T)}{\mathbb{P}(T)}$

But we know that $\mathbb{P}(D \cap T)=\mathbb{P}(T|D)\mathbb{P}(D)$

So $\mathbb{P}(D|T)=\frac{\mathbb{P}(T|D)\mathbb{P}(D) }{\mathbb{P}(T)}$

Now what is $\mathbb{P}(T)$ ?
$T=(T \cap D) \cup (T \cap D^c)$ (because D and its complementary $D^c$ are disjoint)
And we also have $(T \cap D) \cap (T \cap D^c)= \emptyset$

Hence $\mathbb{P}(T)=\mathbb{P}(T \cap D)+\mathbb{P}(T \cap D^c)$
$=\mathbb{P}(T|D)P(D)+\mathbb{P}(T|D^c)P(D^c)$

Therefore, $\boxed{\mathbb{P}(D|T)=\frac{\mathbb{P}(T|D) \mathbb{P}(D)}{\mathbb{P}(T|D)P(D)+\mathbb{P}(T|D^ c)P(D^c)}}
$

3. Originally Posted by Moo
Hello,

So you're looking for P(D|T)

$\mathbb{P}(D|T)=\frac{\mathbb{P}(D \cap T)}{\mathbb{P}(T)}$

But we know that $\mathbb{P}(D \cap T)=\mathbb{P}(T|D)\mathbb{P}(D)$

So $\mathbb{P}(D|T)=\frac{\mathbb{P}(T|D)\mathbb{P}(D) }{\mathbb{P}(T)}$

Now what is $\mathbb{P}(T)$ ?
$T=(T \cap D) \cup (T \cap D^c)$ (because D and its complementary $D^c$ are disjoint)
And we also have $(T \cap D) \cap (T \cap D^c)= \emptyset$

Hence $\mathbb{P}(T)=\mathbb{P}(T \cap D)+\mathbb{P}(T \cap D^c)$
$=\mathbb{P}(T|D)P(D)+\mathbb{P}(T|D^c)P(D^c)$

Therefore, $\boxed{\mathbb{P}(D|T)=\frac{\mathbb{P}(T|D) \mathbb{P}(D)}{\mathbb{P}(T|D)P(D)+\mathbb{P}(T|D^ c)P(D^c)}}
$
Thank you! Thank you so much! Probabilities really confuses me..I don't know what I'll do without your help..but what is the value of P(D^c)? is it equals to 1-P(D) or not? sorry i'm just weak when it comes to probabilities..thank you again!

i went through my notes..i got it..thanks again!