CUET UG 2025 — Mathematics Algebra

Given information:

$$P(E) = 0.25$$P(E)=0.25

$$P(F) = 0.50$$P(F)=0.50

$$P(E \cap F) = 0.14$$P(E∩F)=0.14

The probability that neither E nor F occurs can be found using:

$$P(\text{neither E nor F}) = 1 - P(\text{at least one of E or F})$$P(neither E nor F)=1−P(at least one of E or F)

$$P(\text{neither E nor F}) = 1 - P(E \cup F)$$P(neither E nor F)=1−P(E∪F)

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To find $P(E \cup F)$P(E∪F), use the addition rule:

$$P(E \cup F) = P(E) + P(F) - P(E \cap F)$$P(E∪F)=P(E)+P(F)−P(E∩F)

$$P(E \cup F) = 0.25 + 0.50 - 0.14$$P(E∪F)=0.25+0.50−0.14

$$P(E \cup F) = 0.61$$P(E∪F)=0.61

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Now calculate the probability that neither event occurs:

$$P(\text{neither E nor F}) = 1 - P(E \cup F)$$P(neither E nor F)=1−P(E∪F)

$$P(\text{neither E nor F}) = 1 - 0.61$$P(neither E nor F)=1−0.61

$$P(\text{neither E nor F}) = 0.39$$P(neither E nor F)=0.39

Therefore, the probability that neither E nor F occurs is 0.39.