Question

Medical case histories indicate that different illnesses may produce identical symptoms. Suppose a
particular set of symptoms, which we will denote as event H, occurs only when any one of three
illnesses – A, B, or C – occurs. (Assume that illnesses A, B, and C are mutually exclusive). Studies show
these probabilities of getting the three illnesses: P(A) = 0.01, P(B)=0.005, P(C)= 0.02. The probabilities
of developing the symptoms H, given a specific illness are: P(H|A) = 0.90, P(H|B) = 0.95, P(H|C) = 0.75.

Answer 1

What is probability that a person shows symptoms H?

Assume that an ill person shows the symptoms H. What is the probability that the person has illness A?

Answer 2

i haven't taken stats yet. taking it next month actually :P

Answer 3

do you know who can help me?

Answer 4

This problem can be a little tedious, but the fact that there are few cases makes it tractable.

We have global Mutually Exclusive. That makes it WAY easier.
This is an irrational assumption, but there it is.

P(A) = 0.01, P(B)=0.005, P(C)= 0.02.
Note: P(None) = 1.00 - 0.01 - 0.005 - 0.02 = 0.965

P(H|A) = 0.90, P(H|B) = 0.95, P(H|C) = 0.75.
What we don't have is P(H|None). I guess we should assume it is zero (0.00)
This is an irrational assumption, but there it is.

Add them all up!
P(H|A)*P(A) + P(H|B)*P(B) + P(H|C)*P(C) + P(H|None)*P(None)

Sometimes, there just isn't anything fancy. Jump in and get it ALL done!

Answer 5

could you explain how you got the P(A) P(B) and P(C)

Answer 6

Given in the problem statement. {sigh} You DO have to read it ALL. Yes, it can be a lot to read.

Answer 7

omg i didn't even see that......im so sorry....all i saw was the last few parts...idk why i kept skipping that mart

Answer 8

wait....is that the way to do the first question or the second?

Answer 9

could you help with the second part.......i got 0.009 as the answer

Answer 10

How did you get that. Please show intermediate results.

Answer 11

would it be P(a) times P(HlA)

Answer 12

@tkhunny

Answer 13

thats what I did

Answer 14

Could you help with the other part?

Answer 15

It's in here...

P(H|A)*P(A) + P(H|B)*P(B) + P(H|C)*P(C) + P(H|None)*P(None)

Please provide those four products. Do NOT add them up.

Answer 16

wait....is that the way I answer this question "What is probability that a person shows symptoms H?" or this one "Assume that an ill person shows the symptoms H. What is the probability that the person has illness A?"

Answer 17

because I used that equation for the first question "What is probability that a person shows symptoms H?"

Answer 18

We're adding them up to answer the first part.
We're talking about the individual pieces to answer the second part.

Answer 19

0.009 + 0.00475 + 0.015 + 0

Answer 20

thats what i got

Answer 21

That's right. So, the probability that such symptoms are present is 0.02875. This is the first part.

The second part is asking what percent of 0.02875 is contributed by A, B, and C

A) 0.009/0.02875
B) 0.00475/0.02875
C) 0.015/0.02875

Check to see that it all adds up to one (1).

Answer 22

could you help with another question?

Answer 23

What did you get for that one?

Post on another thread.

Answer 24

It added up to 1 and for A i got 0.313 and ok. btw thank you so much for helping me with this

Answer 25

First: 0.02875

Second:

A) 0.009/0.02875 = 0.3130
B) 0.00475/0.02875 = 0.1652
C) 0.015/0.02875 = 0.5217

Good.

Answer 26

Good day, just wanted to know where u got the 0.009 frm that’s in the numerator … im interested in “Assume that an ill person shows the symptoms H. What is the probability that the person has illness A?”  that part .. thanks @tkhunny