The decimal parts of the logarithms of two numbers taken at random are found to 6 decimal places. What is the probability that the second number can be subtracted from the first number without "borrowing" ?
To find the probability that the second number can be subtracted from the first number without borrowing, we need to determine the conditions that satisfy this condition.
In this case, we know that the decimal parts of the logarithms of the two numbers are given to 6 decimal places. This tells us that we are working with logarithms to the base 10. Let's denote the first number as A and the second number as B.
For A and B to be subtracted without borrowing, each decimal place in the logarithm of A should be greater than or equal to the corresponding decimal place in the logarithm of B.
Now, for a randomly chosen number, the decimal part of its logarithm can range from 0 to 1 because a logarithm represents the exponent to which the base (10 in this case) must be raised to obtain the original number.
To calculate the probability, we'll break it down into separate decimal places:
- For the first decimal place, the probability that A is greater than or equal to B is 1, since the decimal part can range from 0 to 1.
- For the second decimal place, the probability is also 1.
- Similarly, for all six decimal places, the probability is 1.
Since all six decimal places have a probability of 1, the overall probability is the product of the probabilities for each decimal place, which is 1 multiplied by itself six times.
Therefore, the probability that the second number can be subtracted from the first number without borrowing is 1^6 = 1. So, it is certain that the second number can be subtracted from the first number without borrowing.