Assertion -If PA-0- 25- PB-0- 50 and PA- B-0- 14- then the probability that neither A nor B occurs is 0-39- Reason- A- B-A-B

The correct option is C Assertion is correct but Reason is incorrectIf P(A)=0· 25 P(B)=0· 50 P(A∩ B)=0· 14 P(A∪ B) = P(A) +P(B) nbsp;P ...

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Standard IX

Mathematics

Experimental/Empirical Probability

Assertion :If...

Question

Assertion :If $P (A) = 0 \cdot 25, P (B) = 0 \cdot 50$ and $P (A \cap B) = 0 \cdot 14$ , then the probability that neither A nor B occurs is 0.39. Reason: $¯ ¯¯¯¯¯¯¯¯¯¯¯¯ ¯ A \cup B = ¯ ¯¯ ¯ A \cup ¯ ¯¯ ¯ B$

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

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Assertion is correct but Reason is incorrect

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Assertion is incorrect but Reason is correct

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Solution

The correct option is C Assertion is correct but Reason is incorrect
If $P (A) = 0 \cdot 25$
$P (B) = 0 \cdot 50$
$P (A \cap B) = 0 \cdot 14$
$P (A \cup B) = P (A) + P (B) - P (A \cap B)$
$P (A \cup B) = 0.61$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯ ¯ A \cup B = 1 - P (A \cup B) = 0.39$
Hence, assertion is correct but reason is wrong

Suggest Corrections

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Assertion :If P(A)=0⋅25,P(B)=0⋅50 and P(A∩B)=0⋅14, then the probability that neither A nor B occurs is 0.39. Reason: ¯¯¯¯¯¯¯¯¯¯¯¯¯¯A∪B=¯¯¯¯A∪¯¯¯¯B

The correct option is C Assertion is correct but Reason is incorrectIf P(A)=0⋅25P(B)=0⋅50 P(A∩B)=0⋅14P(A∪B)=P(A)+P(B)−P(A∩B)P(A∪B)=0.61¯¯¯¯¯¯¯¯¯¯¯¯¯¯A∪B=1−P(A∪B)=0.39Hence, assertion is correct but reason is wrong

Assertion :If $P (A) = 0 \cdot 25, P (B) = 0 \cdot 50$ and $P (A \cap B) = 0 \cdot 14$ , then the probability that neither A nor B occurs is 0.39. Reason: $¯ ¯¯¯¯¯¯¯¯¯¯¯¯ ¯ A \cup B = ¯ ¯¯ ¯ A \cup ¯ ¯¯ ¯ B$