CBSE 2025, Set 1

Question

**Case Study 3**

Based on 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, $A_3$: people with poor health.

During a pandemic, the chances of people from categories $A_1$, $A_2$ and $A_3$ contracting the disease 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? [2 marks]

(ii) Given that the person has **not** contracted the disease, what is the probability that the person is from category $A_2$? [2 marks]

Edvaya · Accepted Answer

The correct answer is Subjective. Let $D$ = event of contracting the disease. $P(A_1) = 0.7$, $P(A_2) = 0.2$, $P(A_3) = 0.1$ $P(D|A_1) = 0.25$, $P(D|A_2) = 0.35$, $P(D|A_3) = 0.50$ **(i) P(D) by the Total Probability Theorem:** $$P(D) = P(A_1)P(D|A_1) + P(A_2)P(D|A_2) + P(A_3)P(D|A_3)$$ $$= 0.7 	imes 0.25 + 0.2 	imes 0.35 + 0.1 	imes 0.50 = 0.175 + 0.070 + 0.050$$ $$\boxed{P(D) = 0.295 = \frac{59}{200}}$$ **(ii) $P(A_2 \mid D')$ — given person is NOT diseased:** $$P(D') = 1 - 0.295 = 0.705$$ $$P(D'|A_2) = 1 - 0.35 = 0.65$$ $$P(A_2 \cap D') = P(A_2) \cdot P(D'|A_2) = 0.2 	imes 0.65 = 0.13$$ $$P(A_2|D') = \frac{P(A_2 \cap D')}{P(D')} = \frac{0.13}{0.705} = \frac{130}{705} = \boxed{\frac{26}{141}}$$

Edvaya Target

Edvaya Aspire

Take a Free Mock Test

Target Subjects

Aspire Subjects

Subject Mastery

Mathematics > Statistics And Probability > Probability

Keep Practicing

Current Question

Community Stats

Track Your Progress

Edvaya Target

We Value Your Privacy