probability.Please help

**mika** · March 14th 2006, 08:44 PM

There are two diagnostic tests for a disease. Among those who have the disease, 10% give negative results for the first test and, independently of this, 5% give negative results on the second test. Among those who do not have the disease, 80% give negative results on the first test and, independently, 70% give negative results on the second test. From historical data it is known that 20% of those sent by their GPs for diagnostic testing actually have the disease.

a) If both tests are negative, what is the probability that the person tested has the disease?

b) If both tests are positive, what is the probability that the person tested has the disease?

c) If the first test gives a positive result, what is the probability that the second test will also be positive?

**CaptainBlack** · March 22nd 2006, 12:06 AM

Originally Posted by mika

There are two diagnostic tests for a disease. Among those who have the disease, 10% give negative results for the first test and, independently of this, 5% give negative results on the second test. Among those who do not have the disease, 80% give negative results on the first test and, independently, 70% give negative results on the second test. From historical data it is known that 20% of those sent by their GPs for diagnostic testing actually have the disease.

a) If both tests are negative, what is the probability that the person tested has the disease?

Sequential application of Bayes' theorem will do this.

Bayes theorem:

$ p(A|B)=\frac{p(B|A)p(A)}{p(B)} $ ,

which should be read as: "The probability of event A occurring given that
event B has occurred is equal to the probability of B occurring given that A has
occurred times the unconditional probability of A divided by the unconditional
probability of B".

The two stages of this problem are:
i. The person comes in (having probability $p_0$ of 0.2 of being infected), they take
test 1. On the basis of this test they now have a probability $p_1$ of being infected.
ii. The person (now with probability $p_1$ of being infected takes test 2.
On the basis of this test they now have a probability $p_2$ of being infected.

Let's look at stage i:
The test result is negative so:

$ p_1=p(\mbox{infected}|\mbox{neg on t1})=\frac{p(\mbox{neg on t1}|\mbox{infected})p_0}{p(\mbox{neg on t1})} $

$ =\frac{0.1 \times 0.2}{0.1 \times 0.2+0.8 \times 0.8}\approx 0.03 $

(note $p(\mbox{neg on t1})$ is the probability of negative on test 1 if person infected
times the probability that person is infected plus the probability of
negative on test 1 if person not infected times the probability that the person
is not infected).

Now let's look at stage ii:
Now we repeat the process again for the second test (now the probability
that the person is infected is $p_1$ ):

$ p_2=p(\mbox{infected}|\mbox{neg on t2})=\frac{p(\mbox{neg on t2}|\mbox{infected})p_1}{p(\mbox{neg on t2})} $

$ =\frac{0.05 \times 0.03}{0.05 \times 0.03+0.7 \times 0.97}\approx 0.002 $

,

or about $0.2\%$

RonL

**CaptainBlack** · March 22nd 2006, 04:47 AM

Originally Posted by mika

There are two diagnostic tests for a disease. Among those who have the disease, 10% give negative results for the first test and, independently of this, 5% give negative results on the second test. Among those who do not have the disease, 80% give negative results on the first test and, independently, 70% give negative results on the second test. From historical data it is known that 20% of those sent by their GPs for diagnostic testing actually have the disease.

c) If the first test gives a positive result, what is the probability that the second test will also be positive?

If the first test gives a positive result:

$ p(\mbox{infected}|\mbox{pos on t1})=\frac{p(\mbox{pos t1}|\mbox{infected})p_0}{p(\mbox{pos t1})} $

$ =\frac{0.9 \times 0.2}{0.9 \times 0.2+0.2 \times 0.8}\approx 0.529 $

$ p(\mbox{uninfctd}|\mbox{pos on t1})=\frac{p(\mbox{positive on t1}|\mbox{uninfctd})(1-p_0)}{p(\mbox{pos on t1})} $

$ =\frac{0.2 \times 0.8}{0.9 \times 0.2+0.2 \times 0.8}\approx 0.471 $

.

Finally:

$ p(\mbox{pos on t2}|\mbox{pos on t1})=$ $p(\mbox{uninfctd}|\mbox{pos on t1})p(\mbox{pos on t2}|\mbox{uninfctd})$ $+p(\mbox{infected}|\mbox{pos on t1})p(\mbox{pos on t2}|\mbox{infected}) $

$ =0.471 \times 0.3+0.529 \times 0.95\approx 0.644 $

RonL

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