3. Three digits are chosen at random from 1 2 3 4 5 6 7 8 9 Without repeating any digit. What is the probability that the product is odd? A. \( 2 / 3 \) B. \( 7 / 48 \) C. \( 5 / 42 \) D. \( 5 / 108 \)

To determine the probability that the product of three randomly chosen digits from 1 to 9 is odd, we first need to analyze the digits. The odd digits available are 1, 3, 5, 7, and 9, making a total of 5 odd digits. The even digits are 2, 4, 6, and 8, totaling 4 even digits. To achieve an odd product, all three selected digits must be odd. Thus, we need to calculate the total ways to choose 3 digits from 9 and the ways to choose 3 odd digits from 5. Total ways to choose 3 digits from 9: \[ \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \] Total ways to choose 3 odd digits from 5: \[ \binom{5}{3} = \frac{5 \times 4}{2 \times 1} = 10 \] Now, the probability that all three digits are odd is: \[ P(\text{odd product}) = \frac{\text{Ways to choose 3 odd digits}}{\text{Total ways to choose 3 digits}} = \frac{10}{84} = \frac{5}{42} \] So, the answer is \( C. \frac{5}{42} \).