<p>If any four numbers are selected and they are multiplied, then the probability that the last digit will be 1, 3, 5 or 7, is</p>
<p>(a) \(\frac{4}{625}\)</p>
<p>(b) \(\frac{18}{625}\)</p>
<p>(c) \(\frac{16}{625}\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Last digit of product is odd iff all factors are odd; specifically 1,3,5,7 requires excluding 9
<p>The last digit of a product depends only on the last digits of the factors. Last digit is odd (1, 3, 5, or 7) only if all four numbers have odd last digits. Last digits are 0-9 (10 options). Odd last digits: 1, 3, 5, 7, 9 (5 options). P(all four have odd last digits) = (5/10)⁴ = (1/2)⁴ = 1/16. But we want last digit to be specifically 1, 3, 5, or 7 (excluding 9). P(each number ends in 1,3,5,7) = (4/10)⁴ = (2/5)⁴ = 16/625.</p>
Correct Answer: C