CUET UG 2025 — Mathematics Algebra

A die is thrown three times, and the first throw is already 5.

The total sum needed is 14. Since the first throw is 5, the sum needed from the 2nd and 3rd throws is:

$$14 - 5 = 9$$14−5=9

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When throwing a die twice, each die can show 1, 2, 3, 4, 5, or 6.

Total possible outcomes $= 6 \times 6 = 36$=6×6=36

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Finding the outcomes where the 2nd and 3rd throws sum to 9:

2nd throw = 3, 3rd throw = 6 → sum = 9

2nd throw = 4, 3rd throw = 5 → sum = 9

2nd throw = 5, 3rd throw = 4 → sum = 9

2nd throw = 6, 3rd throw = 3 → sum = 9

Favorable outcomes = 4

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The probability is:

$$\text{Probability} = \dfrac{\text{Favorable outcomes}}{\text{Total possible outcomes}}$$Probability=Total possible outcomesFavorable outcomes​

$$\text{Probability} = \dfrac{4}{36}$$Probability=364​

$$\text{Probability} = \dfrac{1}{9}$$Probability=91​

Therefore, the probability of getting a sum of 14 is $\dfrac{1}{9}$91​.