CUET UG 2025 — General Test Quantitative Reasoning

The basket contains:

Red marbles: 4

Blue marbles: 5

Green marbles: 3

Total marbles: $4 + 5 + 3 = 12$4+5+3=12

Three marbles are picked at random.

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"At least one blue" means 1 blue, 2 blues, or 3 blues. Calculating all three cases separately requires multiple calculations.

Using the complement:

$$P(\text{at least one blue}) = 1 - P(\text{no blue marbles})$$P(at least one blue)=1−P(no blue marbles)

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If no blue marbles are picked, all three marbles must be red or green.

Non-blue marbles = $4 + 3 = 7$4+3=7

Total ways to pick 3 marbles from 12:

$$C(12,3) = \dfrac{12 \times 11 \times 10}{3 \times 2 \times 1}$$C(12,3)=3×2×112×11×10​

$$= \dfrac{1320}{6}$$=61320​

$$= 220$$=220

Ways to pick 3 marbles from 7 non-blue marbles:

$$C(7,3) = \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1}$$C(7,3)=3×2×17×6×5​

$$= \dfrac{210}{6}$$=6210​

$$= 35$$=35

Therefore:

$$P(\text{no blue}) = \dfrac{35}{220}$$P(no blue)=22035​

$$= \dfrac{7}{44}$$=447​

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$$P(\text{at least one blue}) = 1 - \dfrac{7}{44}$$P(at least one blue)=1−447​

$$= \dfrac{44}{44} - \dfrac{7}{44}$$=4444​−447​

$$= \dfrac{37}{44}$$=4437​

The answer is $\dfrac{37}{44}$4437​ (Option 4).