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Question 8063 – Engineering-Mathematics
December 10, 2023Logical-Reasoning
December 10, 2023GATE 2014 [Set-3]
| Question 58 |
Let S be a sample space and two mutually exclusive events A and B be such that A∪B = S. If P(∙) denotes the probability of the event, the maximum value of P(A)P(B) is __________.
| 0.25 | |
| 0.26 | |
| 0.27 | |
| 0.28 |
Question 58 Explanation:
We know that
P(A∪B) = P(A) + P(B) + P(A∩B) = 1 →①
But, as A and B are mutually exclusive events
P(A∩B) = 0
∴ P(A∪B) = P(A) + P(B) = 1 →②
Arithmetic mean of two numbers ≥ Geometric mean of those two numbers
(P(A)+P(B))/2 ≥ √(P(A)∙P(B))
1/2 ≥ √(P(A)∙P(B)) (∵from ②)
Squaring on both sides
1/4 ≥ P(A)∙P(B)
P(A)∙P(B) ≤ 1/4
∴ Maximum value of P(A)P(B) = 1/4 = 0.25
P(A∪B) = P(A) + P(B) + P(A∩B) = 1 →①
But, as A and B are mutually exclusive events
P(A∩B) = 0
∴ P(A∪B) = P(A) + P(B) = 1 →②
Arithmetic mean of two numbers ≥ Geometric mean of those two numbers
(P(A)+P(B))/2 ≥ √(P(A)∙P(B))
1/2 ≥ √(P(A)∙P(B)) (∵from ②)
Squaring on both sides
1/4 ≥ P(A)∙P(B)
P(A)∙P(B) ≤ 1/4
∴ Maximum value of P(A)P(B) = 1/4 = 0.25
Correct Answer: A
Question 58 Explanation:
We know that
P(A∪B) = P(A) + P(B) + P(A∩B) = 1 →①
But, as A and B are mutually exclusive events
P(A∩B) = 0
∴ P(A∪B) = P(A) + P(B) = 1 →②
Arithmetic mean of two numbers ≥ Geometric mean of those two numbers
(P(A)+P(B))/2 ≥ √(P(A)∙P(B))
1/2 ≥ √(P(A)∙P(B)) (∵from ②)
Squaring on both sides
1/4 ≥ P(A)∙P(B)
P(A)∙P(B) ≤ 1/4
∴ Maximum value of P(A)P(B) = 1/4 = 0.25
P(A∪B) = P(A) + P(B) + P(A∩B) = 1 →①
But, as A and B are mutually exclusive events
P(A∩B) = 0
∴ P(A∪B) = P(A) + P(B) = 1 →②
Arithmetic mean of two numbers ≥ Geometric mean of those two numbers
(P(A)+P(B))/2 ≥ √(P(A)∙P(B))
1/2 ≥ √(P(A)∙P(B)) (∵from ②)
Squaring on both sides
1/4 ≥ P(A)∙P(B)
P(A)∙P(B) ≤ 1/4
∴ Maximum value of P(A)P(B) = 1/4 = 0.25