<p>A box contains 11 tickets numbered 1 to 11. Two tickets are drawn simultaneously. Let \(E_1\) = sum is even, \(E_2\) = product is even. Which are CORRECT?</p>
Step-by-Step Solution
Key Concept: Odd = {1,3,5,7,9,11} = 6, Even = {2,4,6,8,10} = 5. P(E_1)=25/55=5/11, P(E_2)=40/55=8/11.
<p>$P(E_1) = \frac{C(6,2)+C(5,2)}{C(11,2)} = \frac{25}{55} = \frac{5}{11}$. A is TRUE.</p><p>$P(E_2) = 1-\frac{C(6,2)}{55}=1-\frac{15}{55}=\frac{8}{11}$. B is TRUE.</p><p>$E_1 \subset E_2$?: when sum is even and both are odd, product is odd \to E_1 ⊄ E_2. C: E_1\subsetE_2 is FALSE \to C is TRUE.</p><p>$P(E_1\cap E_2)=P(\text{both even})=\frac{10}{55}=\frac{2}{11}$. $P(E_1)P(E_2)=\frac{5}{11}\cdot\frac{8}{11}=\frac{40}{121}\neq\frac{2}{11}$. Not independent \to D is FALSE.</p><p>Answer: BC ✓</p>
Correct Answer: BC