Question:
If \(E_1\) and \(E_2\) are two events of a sample space such that \(P(E_1)=0.8\), \(P(E_2)=0.7\) and \(P(E_1\cap E_2)\geq c\), then \(c=\)
If \(E_1\) and \(E_2\) are two events of a sample space such that \(P(E_1)=0.8\), \(P(E_2)=0.7\) and \(P(E_1\cap E_2)\geq c\), then \(c=\)
Show Hint
For any two events:
\[
P(A\cap B)\geq P(A)+P(B)-1
\]
This is a very important lower bound formula in probability theory.
Updated On: Jun 17, 2026
- \(0.5\)
- \(0.6\)
- \(0.7\)
- \(0.65\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
For any two events \(A\) and \(B\),
\[
P(A\cup B)=P(A)+P(B)-P(A\cap B)
\]
Also, probability of any event cannot exceed \(1\):
\[
P(A\cup B)\leq 1
\]
Using these two facts, we can determine the minimum possible value of \(P(A\cap B)\).
Step 1: Use the probability addition theorem. Given: \[ P(E_1)=0.8 \] \[ P(E_2)=0.7 \] Using: \[ P(E_1\cup E_2) = P(E_1)+P(E_2)-P(E_1\cap E_2) \] Substituting values: \[ P(E_1\cup E_2) = 0.8+0.7-P(E_1\cap E_2) \] \[ = 1.5-P(E_1\cap E_2) \]
Step 2: Use the fact that probability cannot exceed \(1\). Since: \[ P(E_1\cup E_2)\leq 1 \] Therefore, \[ 1.5-P(E_1\cap E_2)\leq 1 \] Rearranging: \[ -P(E_1\cap E_2)\leq -0.5 \] Multiplying by \(-1\) reverses inequality: \[ P(E_1\cap E_2)\geq 0.5 \] Thus, \[ c=0.5 \] Hence the required answer is: \[ \boxed{0.5} \]
Step 1: Use the probability addition theorem. Given: \[ P(E_1)=0.8 \] \[ P(E_2)=0.7 \] Using: \[ P(E_1\cup E_2) = P(E_1)+P(E_2)-P(E_1\cap E_2) \] Substituting values: \[ P(E_1\cup E_2) = 0.8+0.7-P(E_1\cap E_2) \] \[ = 1.5-P(E_1\cap E_2) \]
Step 2: Use the fact that probability cannot exceed \(1\). Since: \[ P(E_1\cup E_2)\leq 1 \] Therefore, \[ 1.5-P(E_1\cap E_2)\leq 1 \] Rearranging: \[ -P(E_1\cap E_2)\leq -0.5 \] Multiplying by \(-1\) reverses inequality: \[ P(E_1\cap E_2)\geq 0.5 \] Thus, \[ c=0.5 \] Hence the required answer is: \[ \boxed{0.5} \]
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