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 \(c=\)

• A. \(0.5\)
• B. \(0.6\)
• C. \(0.7\)
• D. \(0.65\)

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.

Answer 1

The Correct Option is A

Step 1: Recall the addition rule.
For two events, $P(E_1\cup E_2)=P(E_1)+P(E_2)-P(E_1\cap E_2)$.
Step 2: Put in the 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 3: Use the probability limit.
A probability can never be more than $1$, so $P(E_1\cup E_2)\le1$.
Step 4: Form the inequality.
So $1.5-P(E_1\cap E_2)\le1$.
Step 5: Solve for the intersection.
Rearranging gives $P(E_1\cap E_2)\ge0.5$.
Step 6: Read the value of $c$.
The smallest the intersection can be is $0.5$, so $c=0.5$. \[ \boxed{0.5} \]