CMAT questions on Modern Math - Probability

Total marbles = 1 + 3 + 2 + 4 = 10.

• A: P(yellow) = 4/10, which is (IV).
• B: P(green) = 3/10, which is (I).
• C: P(green or yellow) = (3 + 4)/10 = 7/10, which is (II).
• D: P(red or yellow) = (1 + 4)/10 = 5/10 = 1/2, which is (III).

(The colours are mutually exclusive, so "or" means simply adding the counts.)

Matching: A–IV, B–I, C–II, D–III.

Answer: option 3

Key idea: P(hits at least once) = 1 − P(misses all four shots).

The miss probabilities are 0.9, 0.8, 0.65 and 0.55.

P(all miss) = 0.9 × 0.8 × 0.65 × 0.55 = 0.72 × 0.3575 = 0.2574.

P(hit) = 1 − 0.2574 = 0.7426.

(The player stops after the first hit, but that does not change the answer. Not hitting at all still means missing all four shots.)

Answer: 0.7426

Total marbles = 1 + 3 + 2 + 4 = 10.

• A: P(yellow) = 4/10 = 2/5, not 1/5. False.
• B: P(green) = 3/10. True.
• C: P(yellow or green) = (4 + 3)/10 = 7/10. True.
• D: P(red or yellow) = (1 + 4)/10 = 1/2, not 3/10. False.

Answer: (B) and (C) only (option 2)

Asked: P(2 red and 1 green) when 3 balls are drawn from 15 red and 10 green.

Total ways = \( ^{25}C_3 = \frac{25 \times 24 \times 23}{6} = 2300 \).

Favourable ways = \( ^{15}C_2 \times ^{10}C_1 = 105 \times 10 = 1050 \).

Probability = \( \frac{1050}{2300} = \frac{21}{46} \).

Answer: 21/46

Asked: P(first ball red, then second ball yellow), drawn without replacement.

Total balls = 6 + 11 + 5 = 22.

P(first red) = 6/22.

After one red is removed, 21 balls remain, still with 11 yellow. P(second yellow) = 11/21.

Probability = \( \frac{6}{22} \times \frac{11}{21} = \frac{66}{462} = \frac{1}{7} \).

Answer: 1/7