<p>The probability that a randomly chosen 5-digit number formed by selecting 5 distinct digits from \(\{1, 2, \ldots, 9\}\) is an odd number is</p>
Step-by-Step Solution
Key Concept: From {1,...,9}: 5 odd digits, 4 even. P(last digit is odd) = 5/9.
<p>Total 5-digit numbers from 9 distinct digits (any arrangement of 5 from 9): $P(9,5) = 9 \times 8 \times 7 \times 6 \times 5$.</p><p>For odd number, last digit must be odd: 5 choices from {1,3,5,7,9}. Remaining 4 positions: $P(8,4)$ ways.</p><p>$P(\text{odd}) = \dfrac{5 \times P(8,4)}{P(9,5)} = \dfrac{5}{9}$</p>
Correct Answer: A