Question

Probability,

A test was developed to detect a particular type of arthritis in individuals over 50 years old. From a survey it is known that 10% of individuals in this age group suffer from this type of arthritis. The test was used on individuals confirmed with the disease, and a positive result was obtained 85% of the times. When the test was used on those confirmed without the disease, the negative result was reported 4% of the times.


a) What is the probability that an individual has the disease given that the test is positive?
b) What is the probability that an individual has the disease given that the test is negative?

Answer 1

@ParthKohli @dan815 @Nnesha @nincompoop @zzr0ck3r @inkyvoyd @Somy @calculusxy

Answer 2

@KyanTheDoodle @Skyz @aryandecoolest @Mchilds15 @Gokuporter @Ahsome @ayyookyndall

Answer 3

u know so many people

Answer 4

@Ahsome any idea

Answer 5

:/

Answer 6

@Christos why did u make him ur fan if he is not replying

Answer 7

I thought he would be intelligent

Answer 8

I'd say it's 10% for both

Answer 9

how did you solve that ? @alekos

Answer 10

The probability that an individual has the disease is 10% regardless of whether the test is positive or negative

Answer 11

but you have to think about what it is saying in the question specifically @phi

Answer 12

we use P(a|b) = P(a and b)/P(b)
to use this we need P( + test)
P(+) = P(+|d) + P(+ / not d)

Answer 13

hmm what is P( + test) ?

Answer 14

Probability of getting a positive test
if you have the disease it is 0.85
if you don't have it it is 0.04 (i.e. the test responds postively, even though there is no disease)

Answer 15

so the answer to a is 85% ?

Answer 16

in this situation d is 0.1 ?

Answer 17

my thinking might be muddled here, and it will take a few minutes to clear it up

Answer 18

hmm ok

Answer 19

Ok, here are the gory details


Some theorems
1 Pr(a | b) = Pr (a & b) / Pr(b) from which we get
2 Pr(a & b) = Pr(b) * Pr(a | b)
3 Pr(a) = Pr(b) * Pr(a|b) + Pr(~b) * Pr(a | ~b)

Given
Pr(d)= 0.1
Pr(~d) = 0.9

Pr(+ | d) = 0.85
Pr(+ | ~d) = 0.04

Answer 20

we want to find
Pr(d|+) which by eq 1 is
Pr(d | +) = Pr( d & +) / Pr(+)

which means we have to find both Pr's on the right side because they are not given to us explicitly

Answer 21

By eq 2. Pr(a & b) = Pr(b) * Pr(a | b)
Pr (d & +)= Pr( + & d)
and
Pr( + & d) = Pr(d) * Pr ( + | d)

now we have info for this: Pr(d)= 0.1, Pr(+|d) =0.85 so
Pr(+ & d) = 0.085

Answer 22

btw I am using Pr( + & d) to mean the prob of the intersection of positive result and having the disease.

Answer 23

so far we have
Pr(d | +) = Pr( d & +) / Pr(+)
or
Pr(d | +) = 0.085 / Pr(+)

now we need Pr(+)
we use eq 3 Pr(a) = Pr(b) * Pr(a|b) + Pr(~b) * Pr(a | ~b)
Pr(+) = Pr(d) * Pr(+|d) + Pr(~d) * Pr(+ | ~d)
we have these numbers, so
Pr(+) = 0.1 * 0.85 + 0.9 * 0.04
Pr(+)= 0.121

we get
Pr(d | +) = 0.085 / 0.121 = 0.702 or about 70.2%