Probability
Classical definition of probability
Grade 12
Question:
<p>A mapping is selected at random from the set of all the mappings of the set \(A = \{1, 2, \ldots, n\}\) into itself. Find the probability that the mapping selected is an injection.</p>
<p>\(\dfrac{n!}{n^n}\)</p>
<p>\(\dfrac{(n-1)!}{n^{n-1}}\)</p>
<p>\(\dfrac{1}{n!}\)</p>
<p>\(\dfrac{n!}{n^{n-1}}\)</p>
Step-by-Step Solution
Key Concept: An injection from A to itself is a bijection (permutation). The total number of mappings from an n-element set to itself is n^n, while injections are exactly n! permutations. The probability is therefore n!/n^n.
<p><strong>Step 1:</strong> Count total mappings from A to A. Since each of the n elements can map to any of the n elements independently, total mappings = n^n.</p><p><strong>Step 2:</strong> Count injective mappings. For an injection f: A → A where A is finite, injectivity implies surjectivity (hence bijectivity). For the first element, we have n choices; for the second, (n-1) choices; for the third, (n-2) choices, and so on. Total injections = n(n-1)(n-2)...1 = n!</p><p><strong>Step 3:</strong> Apply probability formula. P(injection) = (Number of injections)/(Total mappings) = n!/n^n</p><p>∴ Answer: <strong>n!/n^n</strong></p>
Correct Answer: B