In: Statistics and Probability
NYHA I |
NYHA II |
NYHA III |
NYHA IV |
Total |
|
Age < 40 years |
60 |
30 |
20 |
10 |
120 |
Age ≥ 40 years |
100 |
80 |
60 |
40 |
280 |
Total |
160 |
110 |
80 |
50 |
400 |
Solution
Back-up Theory
If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then
Conditional Probability of B given A, denoted by P(B/A) = P(B ∩ A)/P(A)……....….(1)
P(A/B) = P(B/A) x {P(A)/P(B)}………………………..………………………..…….(2)
Now, to work out the solution,
Part (a)
Patients classified as Type II, III, and IV at the heart clinic are referred to imaging clinic.
Out of 400 patients at the heart clinic, 240 are classified as Type II, III, and IV.
Combining the above two statements,
probability that a randomly selected patient from the heart clinic will be referred to the imaging clinic
= 240/400
= 0.6 Answer 1
Part (b)
This probability cannot be determined. Answer 2
If we also know how many of the 60 patients the imaging clinic classified as Type IV had been classified as Type IV by the heart clinic, then this probability can be determined.
Part (c)
Now, we know that out of 60 patients the imaging clinic classified as Type IV, 45 had been classified as Type IV by the heart clinic.
Sub-part (i)
If a patient is diagnosed with NYHA IV at the heart clinic, the likely-hood they will be diagnosed as NYHA IV at the imaging clinic
= 45/60
= 0.75 Answer 3
Sub-part (ii)
Just for convenience of presentation and explaining, let A represent the event that a patient is diagnosed with NYHA IV at the imaging clinic and B represent the event a patient is diagnosed with NYHA IV at the heart clinic.
Then, by Answer 3, P(A/B) = 0.75.
What we want to evaluate is: P(B/A), which vide (2)
= P(A/B) x P(B)/P(A)
= 0.75 x (50/400)/(45/240)
= 0.5 Answer 4
DONE