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 of disease contraction in a randomly tested individual

Total population: 1000

\(A_1\): 700 individuals, 25% (0.25) probability of contracting disease.

\(A_2\): 200 individuals, 35% (0.35) probability of contracting disease.

\(A_3\): 100 individuals, 50% (0.50) probability of contracting disease.

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

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

The probability is \(0.295\), equivalent to \(29.5%\).

(ii) Probability of belonging to category \(A_2\) given no disease contraction

Total individuals not contracting disease: \(1000 - 295 = 705\)

Individuals in \(A_2\) with no disease contraction: 200 individuals, probability of not contracting disease = \(1 - 0.35 = 0.65\).

Number of individuals from \(A_2\) not contracting disease: \(200 \times 0.65 = 130\)

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

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