In a random sample of three integers between 1 and 365 inclusive, what

is the probability, to three decimal place accuracy, that
(a) the three
numbers will all be di�fferent;
(b) the three numbers will not all be
di�fferent?

To find the probability in this scenario, we need to calculate the number of favorable outcomes and the total number of possible outcomes.

a) The probability that all three numbers are different:

Favorable outcomes: We need to select three integers from a set of 365 integers without repetition. The first can be any of the 365 numbers, the second can be any of the remaining 364 numbers, and the third can be any of the remaining 363 numbers. So, the number of favorable outcomes is 365 * 364 * 363.

Total outcomes: We have to choose three integers from a set of 365 integers with repetition allowed. This means that for each selection, we have 365 choices. So, the total number of outcomes is 365 * 365 * 365.

Now we can calculate the probability:
P(a) = favorable outcomes / total outcomes
P(a) = (365 * 364 * 363) / (365 * 365 * 365)

b) The probability that not all three numbers are different is equal to 1 - P(a). We can subtract the probability of getting all three different numbers from 1 to find this probability.

P(b) = 1 - P(a)

To get the answer to both parts (a) and (b), we need to calculate the above equations using a calculator or a programming language capable of handling large numbers.