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Consider the experiment of rolling a die. Let A be the event 'getting a prime number', B be the event 'getting an odd number'. However C be the event containing only highest number of die. If P is the probability of the events.

(A) Here, S={1,2,3,4,5,6},A={2,3,5},B={1,3,5},C={6}S = \{1,2,3,4,5,6\}, A = \{2,3,5\}, B = \{1,3,5\}, C = \{6\}

(B) A∪B={1,2,3,5}A \cup B = \{1,2,3,5\} & A∩B={3,5}A \cap B = \{3,5\}

(C) P(A)≥0,P(B)≥0,P(C)≥0P(A) \geq 0, P(B) \geq 0, P(C) \geq 0

(D) P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

Arrange the above events in chronological order and choose the correct answer from the options given below:

Solution

✅ Correct Option: 1

The logical working order is: first write the sample space and the sets A, B and C, which is (A); then form their union and intersection, which is (B); then note that each probability is non negative, which is (C); and finally apply the addition rule for P(A∪B)P(A \cup B), which is (D). Hence (A), (B), (C), (D).

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