We have three numerals and two letters here. Since there are in theory a choice of 26 letters and there are 26^2=676 combinations from AA to ZZ, there are only 26 characters combinations of repeated letters AA, BB, CC, etc. 26 out of 676=26. With the numerals, if the leading numerals are a and b and the last numeral is c, we have pairs ab, ac and bc. Now assuming all numerals are valid we have 00 to 99 for each of ab, ac and bc, that is 100 for each. Within the 100 we have 10 repeated numerals: 00, 11, 22, etc. so there are 10 out of 100. When we combine probabilities with "or" rather than "and", we add the individual probabilities. So it's 1/26 for the letters, plus 1/10 for each pair of numerals=3/10. That makes 1/26+3/10=(5+39)/130=44/130=22/65.