If P(A) = (1)/(3) , P(B) = (1)/(2) and P(A B) = (1)/(4) , then what is the value of P(A^c B^c) ?

Correct answer: B. (5)/(12) Using the formula for the probability of the union of two events: P(A B) = P(A) + P(B) - P(A B) Substituting the given values: P(A B) = (1)/(3) + (1)/(2) - (1)/(4) P(A B) = (4 + 6 - 3)/(12) = (7)/(12) By De Morgan's laws, A^c B^c = (A B)^c . Therefore, the probability is: P(A^c B^c) = 1 - P(A B) P(A^c B^c) = 1 - (7)/(12) = (5)/(12)

NDA 2025

Exam: NDA
Year: 2025
Subject: Mathematics
Chapter: Probability

NDA 2025 · Mathematics · Probability

If P(A) = (1)/(3) , P(B) = (1)/(2) and P(A B) = (1)/(4) , then what...

P(A) = \dfrac{1}{3}

P(B) = \dfrac{1}{2}

and

P(A \cap B) = \dfrac{1}{4}

, then what is the value of

P(A^c \cap B^c)

Options

A $\dfrac{1}{4}$
B $\dfrac{5}{12}$
C $\dfrac{7}{12}$
D $\dfrac{11}{12}$

Correct answer

B. $\dfrac{5}{12}$

Step-by-step solution

Using the formula for the probability of the union of two events:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

Substituting the given values:

P(A \cup B) = \dfrac{1}{3} + \dfrac{1}{2} - \dfrac{1}{4}

P(A \cup B) = \dfrac{4 + 6 - 3}{12} = \dfrac{7}{12}

By De Morgan's laws,

A^c \cap B^c = (A \cup B)^c

.

Therefore, the probability is:

P(A^c \cap B^c) = 1 - P(A \cup B)

P(A^c \cap B^c) = 1 - \dfrac{7}{12} = \dfrac{5}{12}

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