7 out of 20 elk have been caught, tagged, and released
with another 13 of the 20 still untagged.
The probability that exactly 2 of another 5 captured elk
are already tagged would then be expressed as:
[5! ÷ 2! ÷ (5 − 2)!] times (7/20)2 times (13/20)3.
This amounts to 10 times (49/400) times (2197/8000)
or (1076530/3200000) or 0.336415625.