Probability
Independent Events and Subset Selection
Grade 12
Question:
<p>(B) The number of elements in <i>P</i> is more than that in <i>Q</i> is</p>
<p>(P) \(\frac{\binom{2n}{n}}{4^n}\)</p>
<p>(Q) \(\frac{2^{2n} - \binom{2n}{n}}{2^{2n+1}}\)</p>
<p>(R) \(\frac{\binom{2n}{n+1}}{4^n}\)</p>
<p>(S) \(\left(\frac{3}{4}\right)^n\)</p>
<p>(T) \(\frac{\binom{2n}{n}}{4^{n-1}}\)</p>
Step-by-Step Solution
Key Concept: Use symmetry: P(|P| > |Q|) = P(|Q| > |P|), and these two plus P(|P| = |Q|) must sum to 1.
<p>By symmetry, the probability that <i>P</i> has more elements than <i>Q</i> equals the probability that <i>Q</i> has more elements than <i>P</i>. Let these equal probabilities be $x$.</p><p>We have: $x + \frac{\binom{2n}{n}}{4^n} + x = 1$</p><p>Solving: $2x = 1 - \frac{\binom{2n}{n}}{4^n} = \frac{4^n - \binom{2n}{n}}{4^n}$</p><p>Therefore: $x = \frac{2^{2n} - \binom{2n}{n}}{2^{2n+1}}$</p>
Correct Answer: Q