Multiple choice

If four digits from $\left {1, 2, 3, ..... 9\right }$ are taken at random and multiplied together, then the chance that the last digit in the product be $1, 3, 7$ or $9$ is

  1. $\displaystyle \frac{9}{^9P_{4}}$
  2. $\displaystyle \frac{1}{10}$
  3. $\displaystyle \frac{1}{^9C_{4}}$
  4. $\left (\dfrac {4}{9}\right)^4$
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D Correct answer
AI explanation

For the product of four digits to end in 1, 3, 7, or 9, every single digit chosen must be one of those numbers, because any even number or 5 would change the last digit of the product. From the set of 1 through 9, there are exactly 4 choices out of 9 that satisfy this condition. The probability of selecting one of these digits is 4/9, and because the selections are independent, we multiply this probability for each of the four digits. The calculation is 4/9 multiplied by itself four times. The result is (4/9)^4.

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