Probability
Classical Probability
Grade 12
Question:
<p>Let <i>A</i> be a set containing <i>n</i> elements. A subset <i>P</i> of the set <i>A</i> is chosen at random. The set <i>A</i> is reconstructed by replacing the elements of <i>P</i>, and another subset <i>Q</i> of <i>A</i> is chosen at random. The probability that \(P \cap Q\) contains exactly <i>m</i> (<i>m</i> < <i>n</i>) elements is</p>
<p>(1) \(\dfrac{3^{n-m}}{4^n}\)</p>
<p>(2) \(\dfrac{{}^nC_m \cdot 3^m}{4^n}\)</p>
<p>(3) \(\dfrac{{}^nC_m \cdot 3^{n-m}}{4^n}\)</p>
<p>(4) None of these</p>
Step-by-Step Solution
Key Concept: Each element of A independently has probability 1/4 of being in both P and Q (must be chosen in both random selections). Use binomial probability to find exactly m elements in P ∩ Q.
<p><strong>Step 1:</strong> For each element of A, determine the probability it belongs to P ∩ Q.</p><p>An element is in P ∩ Q if and only if it is selected in both P and Q.</p><p>• Probability element is in P = 1/2</p><p>• Probability element is in Q = 1/2</p><p>• Since P and Q are chosen independently: P(element in both) = 1/2 × 1/2 = 1/4</p><p><strong>Step 2:</strong> Model this as a binomial distribution problem.</p><p>We have n elements, each independently in P ∩ Q with probability 1/4. We want exactly m elements in P ∩ Q.</p><p>Number of ways to choose which m elements are in P ∩ Q: C(n,m)</p><p>Probability all m chosen elements are in P ∩ Q: (1/4)^m</p><p>Probability remaining (n-m) elements are NOT in P ∩ Q: (3/4)^(n-m)</p><p><strong>Step 3:</strong> Apply binomial probability formula.</p><p>P(|P ∩ Q| = m) = C(n,m) × (1/4)^m × (3/4)^(n-m)</p><p>This can also be written as: <strong>C(n,m) × 3^(n-m) / 4^n</strong></p><p>∴ Answer: C(n,m) · (1/4)^m · (3/4)^(n-m) or C(n,m) · 3^(n-m) / 4^n</p>
Correct Answer: 3