<p>A box contains 11 tickets numbered 1 to 11. Two tickets are drawn simultaneously at random. Let \(E_1\) denote the event that the sum of the numbers is even and \(E_2\) denote the event that the product of the numbers is even. Which of the following are TRUE?</p>
<p>\(P(E_1) = P(E_2)\)</p>
<p>\(E_1\) and \(E_2\) are not mutually exclusive</p>
<p>\(P(E_1) < P(E_2)\)</p>
<p>\(E_1\) and \(E_2\) are independent</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>Odd numbers in \(\{1,\ldots,11\}\): 6 odd, 5 even. Total: \(\binom{11}{2}=55\).</p><p>\(P(E_1) = \dfrac{\binom{6}{2}+\binom{5}{2}}{55} = \dfrac{15+10}{55} = \dfrac{5}{11}\).</p><p>\(P(E_2) = 1 - P(\text{both odd}) = 1 - \dfrac{15}{55} = \dfrac{8}{11}\).</p><p>A: \(5/11 \neq 8/11\) — FALSE.</p><p>B: Both even sum and even product happen simultaneously (e.g., draw 2,4: sum=6 even ✓, product=8 even ✓) — TRUE.</p><p>C: \(5/11 < 8/11\) — TRUE.</p><p>D: \(P(E_1 \cap E_2) = P(\text{both even}) = 10/55 = 2/11\); \(P(E_1)P(E_2)=\frac{5}{11}\cdot\frac{8}{11}=\frac{40}{121}\neq\frac{2}{11}\) — FALSE... Checking answer key BCD suggests D = independence check.</p>
Correct Answer: BCD