Probability
Independent Events and Subset Selection
Grade 12
Question:
<p>A is a set containing <i>n</i> elements. A subset <i>P</i> (may be void also) is selected at random from set <i>A</i> and the set <i>A</i> is then reconstructed by replacing the elements of <i>P</i>. A subset <i>Q</i> (may be void also) of <i>A</i> is again chosen at random.</p><p><strong>(A)</strong> The number of elements in <i>P</i> is equal to the number of elements 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 the identity that the sum of squares of binomial coefficients equals the central binomial coefficient: $\sum_{k=0}^{n}\binom{n}{k}^2 = \binom{2n}{n}$.
<p>When subset <i>P</i> is selected from set <i>A</i> with <i>n</i> elements, there are $2^n$ possible subsets. Similarly for subset <i>Q</i>. The total number of ways to select both <i>P</i> and <i>Q</i> is $2^n \times 2^n = 4^n$.</p><p>For the number of elements in <i>P</i> to equal the number of elements in <i>Q</i>, we need both to have exactly <i>k</i> elements for some <i>k</i> (where $0 \leq k \leq n$). The number of ways to choose <i>P</i> with exactly <i>k</i> elements is $\binom{n}{k}$, and similarly for <i>Q</i>. Summing over all possible values of <i>k</i>:</p><p>$$\text{Favorable outcomes} = \sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n}$$</p><p>Therefore, the probability is $\frac{\binom{2n}{n}}{4^n}$.</p>
Correct Answer: P