Match probability lists

CMAT 2021 · Quantitative Techniques & Data Interpretation · Medium · Probability

Match List I with List II. Let A and B be events with P(A)=23,P(B)=12P(A) = \frac{2}{3}, P(B) = \frac{1}{2} and P(A∩B)=13P(A \cap B) = \frac{1}{3}

List I: (A) P(A∩Bc)P(A \cap B^c) (B) P(A∪Bc)P(A \cup B^c) (C) P(Ac∩Bc)P(A^c \cap B^c) (D) P(Ac∪Bc)P(A^c \cup B^c)

List II: (I) 23\frac{2}{3} (II) 13\frac{1}{3} (III) 56\frac{5}{6} (IV) 16\frac{1}{6}

Choose the correct answer from the options given below:

  1. A.

    (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

  2. B.

    (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

  3. C.

    (A)-(I), (B)-(III), (C)-(IV), (D)-(II)

  4. D.

    (A)-(II), (B)-(IV), (C)-(III), (D)-(I)

Answer

A

Explanation

  • (A) P(A∩Bc)=P(A)−P(A∩B)=23−13=13  ⟹  P(A \cap B^c) = P(A) - P(A \cap B) = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \implies (II)
  • (B) P(A∪Bc)=P(A)+P(Bc)−P(A∩Bc)=23+12−13=56  ⟹  P(A \cup B^c) = P(A) + P(B^c) - P(A \cap B^c) = \frac{2}{3} + \frac{1}{2} - \frac{1}{3} = \frac{5}{6} \implies (III)
  • (C) P(Ac∩Bc)=1−P(A∪B)=1−(23+12−13)=1−56=16  ⟹  P(A^c \cap B^c) = 1 - P(A \cup B) = 1 - \left(\frac{2}{3} + \frac{1}{2} - \frac{1}{3}\right) = 1 - \frac{5}{6} = \frac{1}{6} \implies (IV)
  • (D) P(Ac∪Bc)=1−P(A∩B)=1−13=23  ⟹  P(A^c \cup B^c) = 1 - P(A \cap B) = 1 - \frac{1}{3} = \frac{2}{3} \implies (I)

Correct match: (A)-(II), (B)-(III), (C)-(IV), (D)-(I).

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