CBSE Class 12 Mathematics Probability Case Studies Set 02

Answer: (d) 0.9

Explanation:
1. Let E₁ = person has COVID, A = person tests positive.
2. 90% of people having COVID are detected, so P(A | E₁) = 90/100 = 0.9.

Teacher's Note:
1. Read ‘given that’ carefully: the condition (has COVID) goes after the bar.
2. 0.1 is the probability of a missed case, P(negative | COVID), not of a positive test.

Answer: (a) 0.01

Explanation:
1. Let E₂ = person does not have COVID.
2. 1% of COVID-free people are wrongly shown positive, so P(A | E₂) = 1/100 = 0.01.

Teacher's Note:
1. This is the false-positive rate; 0.99 is the probability of a correct negative result.
2. P(A | E₂) and P(not A | E₂) always add up to 1: 0.01 + 0.99 = 1.

Answer: (b) 0.999

Explanation:
1. P(E₁) = 0.1% = 0.001
2. P(E₂) = 1 − P(E₁) = 1 − 0.001 = 0.999

Teacher's Note:
1. Convert percentages carefully: 0.1% = 0.1/100 = 0.001, not 0.1.
2. E₁ and E₂ are complementary, so their probabilities must add to 1.

Answer: (c) 0.083

Explanation:
1. By Bayes' theorem, P(E₁ | A) = \( \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} \)
2. = \( \frac{0.001 \times 0.9}{0.001 \times 0.9 + 0.999 \times 0.01} = \frac{0.0009}{0.0009 + 0.00999} \)
3. = \( \frac{0.0009}{0.01089} \) = \( \frac{90}{1089} \) ≈ 0.083

Teacher's Note:
1. Even a fairly accurate test gives few true positives when the disease is rare, because false positives from the large healthy group dominate.
2. Multiplying numerator and denominator by 100000 (90/1089) avoids errors with many decimal places.

Answer: (b) 0.01089

Explanation:
1. By the theorem of total probability, P(A) = P(E₁)P(A | E₁) + P(E₂)P(A | E₂)
2. = 0.001 × 0.9 + 0.999 × 0.01 = 0.0009 + 0.00999
3. P(A) = 0.01089

Teacher's Note:
1. This total P(A) is exactly the denominator used in Bayes' theorem in the previous part.
2. Line up decimal places when adding 0.0009 and 0.00999; misplacing one gives the trap 0.1089.