Question

Based upon the results of regular medical check-ups in a hospital, it was found that out of 1000 people, 700 were very healthy, 200 maintained average health and 100 had a poor health record.
Let \( A_1 \): People with good health,
\( A_2 \): People with average health,
and \( A_3 \): People with poor health.
During a pandemic, the data expressed that the chances of people contracting the disease from category \( A_1, A_2 \) and \( A_3 \) are 25%, 35% and 50%, respectively.
Based upon the above information, answer the following questions:
(i) A person was tested randomly. What is the probability that he/she has contracted the disease?}
(ii) Given that the person has not contracted the disease, what is the probability that the person is from category \( A_2 \)?

Hint:

For these probability problems, always identify the given probabilities clearly, and make use of Bayes' Theorem and the law of total probability when dealing with conditional probabilities and multiple categories.

Answer 1

(i) Probability that a randomly tested person has contracted the disease

Total people = 1000

\(A_1\): 700 people, chance of contracting disease = 25% = 0.25

\(A_2\): 200 people, chance of contracting disease = 35% = 0.35

\(A_3\): 100 people, chance of contracting disease = 50% = 0.50

Probability of contracting disease = \(P(D) = (700 \times 0.25) + (200 \times 0.35) + (100 \times 0.50)\)

\(P(D) = 175 + 70 + 50 = 295 / 1000 = 0.295\)

So, the probability is \(0.295\) or \(29.5%\).

(ii) Probability that the person is from category \(A_2\) given they have not contracted the disease

Total who did not contract disease = \(1000 - 295 = 705\)

People from \(A_2\) = 200, chance of not contracting = \(1 - 0.35 = 0.65\)

Number from \(A_2\) who did not contract = \(200 \times 0.65 = 130\)

Probability = \(P(A_2 | D') = 130 / 705 \approx 0.1844\)

So, the probability is approximately \(0.1844\) or \(18.44%\).