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

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For "at least one blue" problems, use the complement:

P(at least one blue) = 1 - P(no blue marbles)

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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

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Ways to pick 3 marbles with no blue (only red and green):

Non-blue marbles = 4 + 3 = 7

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

$$= \dfrac{210}{6}$$=6210​

$$= 35$$=35

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P(no blue marbles) = $\dfrac{35}{220} = \dfrac{7}{44}$22035​=447​

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

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

$$= \dfrac{37}{44}$$=4437​

Therefore, the probability that at least one marble is blue is $\dfrac{37}{44}$4437​.