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Question:
If A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. then, I. P(not A) = 0.58 II. P(not B) = 0.52 III. P(A or B) = 0.47
If A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. then, I. P(not A) = 0.58 II. P(not B) = 0.52 III. P(A or B) = 0.47
Updated On: Apr 28, 2024
- Only I and II are correct.
- Only II and III are correct.
- Only I and III are true.
- All three statements are correct.
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The Correct Option is A
Solution and Explanation
$I. \,P(not\, A) = 1 - 0.42 = 0.58$
$II.\, P(not\, B) = 1 - P(B) = 1 - 0.48 = 0.52$
$III.\, P(A \,or\, B) = P(A\, \cup \,B) = P(A) + P(B) - P(A \,\cap \,B) = 0.42 + 0.48 - 0.16 = 0.74$
$II.\, P(not\, B) = 1 - P(B) = 1 - 0.48 = 0.52$
$III.\, P(A \,or\, B) = P(A\, \cup \,B) = P(A) + P(B) - P(A \,\cap \,B) = 0.42 + 0.48 - 0.16 = 0.74$
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