Question:
For two events \(A\) and \(B\), a true statement among the following is
For two events \(A\) and \(B\), a true statement among the following is
Show Hint
Remember De Morgan’s law:
\[
\overline{A\cap B}=\bar{A}\cup \bar{B}
\]
and the conditional probability identity:
\[
P(A\cap B)=P(A)P(B|A)
\]
Updated On: Jun 22, 2026
- \[ P(\bar{A}\cup \bar{B}) = 1-P(A)P\left(\frac{B}{A}\right) \]
- \[ P(\bar{A}\cup \bar{B}) = 1-P(A\cup B) \]
- \[ P(\bar{A}\cup \bar{B}) = P(A\cup B) \]
- \[ P(\bar{A}\cup \bar{B}) = P(\bar{A})+P(\bar{B}) \]
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Use De Morgan’s law.
We know that \[ \bar{A}\cup \bar{B} = \overline{A\cap B} \] Therefore, \[ P(\bar{A}\cup \bar{B}) = P(\overline{A\cap B}) \] Using complement rule, \[ P(\overline{A\cap B}) = 1-P(A\cap B) \] Hence, \[ P(\bar{A}\cup \bar{B}) = 1-P(A\cap B) \]
Step 2: Use conditional probability formula.
We know that \[ P(A\cap B) = P(A)\,P(B|A) \] Therefore, \[ P(\bar{A}\cup \bar{B}) = 1-P(A)P(B|A) \] This matches option (1).
Step 3: Verify the remaining options.
Option (2): \[ P(\bar{A}\cup \bar{B}) = 1-P(A\cup B) \] This is incorrect because \[ 1-P(A\cup B)=P(\bar{A}\cap \bar{B}) \] not \[ P(\bar{A}\cup \bar{B}) \] Option (3): \[ P(\bar{A}\cup \bar{B}) = P(A\cup B) \] This is not true in general.
Option (4): \[ P(\bar{A}\cup \bar{B}) = P(\bar{A})+P(\bar{B}) \] This is incorrect because for unions, \[ P(X\cup Y)=P(X)+P(Y)-P(X\cap Y) \]
Step 4: Final conclusion.
Hence, the true statement is \[ \boxed{ P(\bar{A}\cup \bar{B}) = 1-P(A)P(B|A) } \] which corresponds to option (1).
We know that \[ \bar{A}\cup \bar{B} = \overline{A\cap B} \] Therefore, \[ P(\bar{A}\cup \bar{B}) = P(\overline{A\cap B}) \] Using complement rule, \[ P(\overline{A\cap B}) = 1-P(A\cap B) \] Hence, \[ P(\bar{A}\cup \bar{B}) = 1-P(A\cap B) \]
Step 2: Use conditional probability formula.
We know that \[ P(A\cap B) = P(A)\,P(B|A) \] Therefore, \[ P(\bar{A}\cup \bar{B}) = 1-P(A)P(B|A) \] This matches option (1).
Step 3: Verify the remaining options.
Option (2): \[ P(\bar{A}\cup \bar{B}) = 1-P(A\cup B) \] This is incorrect because \[ 1-P(A\cup B)=P(\bar{A}\cap \bar{B}) \] not \[ P(\bar{A}\cup \bar{B}) \] Option (3): \[ P(\bar{A}\cup \bar{B}) = P(A\cup B) \] This is not true in general.
Option (4): \[ P(\bar{A}\cup \bar{B}) = P(\bar{A})+P(\bar{B}) \] This is incorrect because for unions, \[ P(X\cup Y)=P(X)+P(Y)-P(X\cap Y) \]
Step 4: Final conclusion.
Hence, the true statement is \[ \boxed{ P(\bar{A}\cup \bar{B}) = 1-P(A)P(B|A) } \] which corresponds to option (1).
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