Find the number of ways to choose 5 odd numbers from the first 25 numbers (1, 2, 3, 4, 5 … 25) such that probability of selecting a prime from the chosen numbers is 2/5?</br>Note: Put the answer as an integer without any padded zeroes or decimal points. For example, if the answer is 1, then please put 1 as the answer and not 1.0 or 01 or 001.

To solve this problem, we first need to identify the odd prime numbers and the odd non-prime numbers within the first 25 numbers.

The odd prime numbers less than or equal to 25 are: 3, 5, 7, 11, 13, 17, 19, 23. So, there are 8 odd prime numbers.

The odd non-prime numbers less than or equal to 25 are: 1, 9, 15, 21, 25. So, there are 5 odd non-prime numbers.

We want to choose 5 odd numbers such that the probability of selecting a prime from the chosen numbers is 2/5. This means that out of the 5 numbers chosen, 2 must be prime and 3 must be non-prime.

The number of ways to choose 2 primes from 8 is given by the combination formula C(n, r) = n! / [(n-r)!r!], where n is the total number of items, and r is the number of items to choose. So, C(8, 2) = 8! / [(8-2)!2!] = 28.

Similarly, the number of ways to choose 3 non-primes from 5 is C(5, 3) = 5! / [(5-3)!3!] = 10.

Therefore, the total number of ways to choose 5 odd numbers such that the probability of selecting a prime from the chosen numbers is 2/5 is 28 * 10 = 280.