Question

Two natural numbers are selected at random, find the probability that t by 10.

Answer 1

27/100

Answer 2

Solution Explanation:
For the product of two natural numbers to be divisible by 10, it must have both a factor 2 and a factor 5. In terms of natural density,
• Probability that a randomly chosen number is even (i.e. has a factor 2) = ½.
• Probability that a number is divisible by 5 = 1⁄5.

However, note that a number divisible by 10 contributes both factors. A more careful approach is to calculate the complementary probability.

1. Define:
    – P(no even) = probability a number is odd = ½. So for two numbers, P(both odd) = (½)² = ¼.
    – P(no factor 5) = probability a number is not divisible by 5 = 4⁄5. So for two numbers, P(neither divisible by 5) = (4⁄5)² = 16⁄25.

2. For a number to have neither an even factor nor a factor 5, note that an odd number among the naturals appears with probability ½. Out of the odds, the proportion not divisible by 5 is 1 – (1⁄5)/(½) = 1 – (2⁄5) = 3⁄5. (Indeed, since probability a number is odd and divisible by 5 is 1⁄10, the probability a number is odd and not divisible by 5 is ½ – 1⁄10 = 2⁄5.)
    Thus, P(neither even nor divisible by 5) = 2⁄5 for one number, and for two numbers it is (2⁄5)² = 4⁄25.

3. Now using inclusion–exclusion, the probability that the product is NOT divisible by 10 is:

  P(No even or No 5 factor) = P(both odd) + P(neither divisible by 5) – P(neither even nor divisible by 5)
   = ¼ + 16⁄25 – 4⁄25
   = ¼ + 12⁄25.

Convert ¼ to a fraction with denominator 100: ¼ = 25⁄100, and 12⁄25 = 48⁄100.
  So, total = (25 + 48)⁄100 = 73⁄100.

4. Therefore, the probability that the product IS divisible by 10 is:

  1 – 73⁄100 = 27⁄100.