Question:
If \(A\) and \(B\) are two independent events such that \(P(A)=0.3\), \(P(B)=x\), and \(P(A\cup B)=0.44\), then \(x=\)
If \(A\) and \(B\) are two independent events such that \(P(A)=0.3\), \(P(B)=x\), and \(P(A\cup B)=0.44\), then \(x=\)
Show Hint
For independent events,
\[
P(A\cap B)=P(A)P(B)
\]
and for union,
\[
P(A\cup B)=P(A)+P(B)-P(A\cap B)
\]
Updated On: Jun 26, 2026
- \(0.1\)
- \(0.4\)
- \(0.3\)
- \(0.2\)
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The Correct Option is D
Solution and Explanation
Step 1: Use the formula for union of two events.
For any two events \(A\) and \(B\), \[ P(A\cup B)=P(A)+P(B)-P(A\cap B) \]
Step 2: Use the condition of independence.
Since \(A\) and \(B\) are independent events, \[ P(A\cap B)=P(A)P(B) \] Given, \[ P(A)=0.3 \] and \[ P(B)=x \] So, \[ P(A\cap B)=0.3x \]
Step 3: Substitute the values.
Given, \[ P(A\cup B)=0.44 \] Therefore, \[ 0.44=0.3+x-0.3x \]
Step 4: Solve for \(x\).
\[ 0.44=0.3+0.7x \] \[ 0.44-0.3=0.7x \] \[ 0.14=0.7x \] \[ x=\frac{0.14}{0.7} \] \[ x=0.2 \]
Step 5: Final conclusion.
Therefore, \[ \boxed{0.2} \]
For any two events \(A\) and \(B\), \[ P(A\cup B)=P(A)+P(B)-P(A\cap B) \]
Step 2: Use the condition of independence.
Since \(A\) and \(B\) are independent events, \[ P(A\cap B)=P(A)P(B) \] Given, \[ P(A)=0.3 \] and \[ P(B)=x \] So, \[ P(A\cap B)=0.3x \]
Step 3: Substitute the values.
Given, \[ P(A\cup B)=0.44 \] Therefore, \[ 0.44=0.3+x-0.3x \]
Step 4: Solve for \(x\).
\[ 0.44=0.3+0.7x \] \[ 0.44-0.3=0.7x \] \[ 0.14=0.7x \] \[ x=\frac{0.14}{0.7} \] \[ x=0.2 \]
Step 5: Final conclusion.
Therefore, \[ \boxed{0.2} \]
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