CUET UG 2025 — General Test Quantitative Reasoning

When two dice are thrown simultaneously, we need to find the probability that at least one die shows a 4.

"At least one" means one 4 or two 4s (i.e., not zero 4s).

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When throwing 2 dice:

First die has 6 possible outcomes: 1, 2, 3, 4, 5, 6

Second die has 6 possible outcomes: 1, 2, 3, 4, 5, 6

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

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Using the complement method, we find the probability of not getting any 4, then subtract from 1.

For the first die to not show 4:

Possible outcomes: 1, 2, 3, 5, 6 (5 outcomes)

$$P(\text{no 4 on first die}) = \dfrac{5}{6}$$P(no 4 on first die)=65​

For the second die to not show 4:

$$P(\text{no 4 on second die}) = \dfrac{5}{6}$$P(no 4 on second die)=65​

For both dice to not show 4:

$$P(\text{no 4 on either die}) = \dfrac{5}{6} \times \dfrac{5}{6}$$P(no 4 on either die)=65​×65​

$$= \dfrac{25}{36}$$=3625​

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$$P(\text{at least one 4}) = 1 - P(\text{no 4 on either die})$$P(at least one 4)=1−P(no 4 on either die)

$$= 1 - \dfrac{25}{36}$$=1−3625​

$$= \dfrac{36}{36} - \dfrac{25}{36}$$=3636​−3625​

$$= \dfrac{11}{36}$$=3611​

Therefore, the probability that 4 will come up on at least one die is $\dfrac{11}{36}$3611​.