Help with probability S1 question?

Question

A medical test for a specific infection is not fully reliable: if an individual has the infection there is a probability of 0.95 that the test will prove positive, and if an individual doesn't have the infection there is a probability of 0.1 that the test will prove positive. In a certain population the probability that an individual chosen at random will have the infection is 0.2

a) Create a tree diagram to show this information (this i completed and can describe if required)

b) Calculate the probability that if an individual is chosen at random and is tested the result will be negative (i calculated this to be 0.73 as (0.2 x 0.05) + (0.8 x 0.9) = 0.73

c) 10,000 randomly chosen individuals are tested. Show that 2,700 altogether will be expected to have a positive result. Calculate an estimate of the proportion of these individuals with positive test results that are actually infected.

It's question c) i'm stuck with and don't know how to answer. Anyone who could help me out here?

M3 · Accepted Answer

10,000 total
0.2*10,000 = 2000 infected of which 95% = 1900 test +
8,000 healthy out of which 10% = 800 test +

# testing + = 1900 + 800 = 2700
proportion of +'s actually infected =1900/2700 =19/27, ≈ 70.37%

ignoramus · Answer

It often helps to draw up a table like the following :

. . . . . . . . . . . . . .Positive . . . . . . . . .Negative. . . . . . Row
. . . . . . . . . . . . . . Test . . . . . . . . . . . . .Test. . . . . . . .Total
. . . . . . . . . . . . .------------ . . . . . . . .------------- . . . .----------

. Diseased . . . . 0.99 x 0.2. . . . . . . . 0.05 x 0.1

. . . . . . . . . . . . . . = 0.19. . . . . . . . . . . = 0.01. . . . . . . 0.20

. Not diseased. . 0.1 x 0.8 . . . . . . . . .0.9 x 0.8

. . . . . . . . . . . . . . = 0.08 . . . . . . . . . . .= 0.72 . . . . . . .0.80
. . . . . . . . . . . . . ------------ . . . . . . . .-------------

Column total. . . . . .0.27. . . . . . . . . . . . 0.73. . . . . . . .1.00

From which it should be easy to see that the probability that a random individual tests positive = 0.27.  (so out of 10000 individuals, 10000 x 0.27 =2700 test positive).

And the probability that such an individual is really infected = 0.19 / 0.27 = 0.7037, which means that 2700 x 0.7037 = 1900 of them are really infected.

{ And if the bottom right corner of the table above does not add up to 1.00, it is a good indication that some of your calculations are incorrect. (Of course, even if it does total 1.00, it does not guarantee that they ARE correct !) ]

Moonferret · Answer

You know that the possibility of a positive result is 1-0.73=0.27
This can also be obtained by (0.2*0.95)+(0.8*0.1)=0.27

0.27*10,000=2,700 as required

The 0.2*0.95 represents the "actually infected positives", so 19/27 of the total are in this category.

Elizabeth M · Answer

c) P(-)=1-P(+)=0.27. From a random 27000 individuals each with P(+)=0.27 the 
number X with + has a binomial distribution with n=10000 and p=0.27, the expected
number E(X)=np=2700.
Let I=person is infected, += test result positive, P(I|+) = person is infected
given that the test is +. Then P(I|+) is required = P(+|I)P(I)/P(+)
= 0.95X0.2/0.27 = ??

ridings · Answer

one million) 2) unquestionably, the uncomplicated thank you to do all of us of those is to be attentive to what the symbols mean after which comedian strip a Venn diagram, filling in all the numbers given. 3) Is binomial distribution 4) in case you could no longer visualise it, draw a tree diagram of all paths required.

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Help with probability S1 question?

5 Answers