Probability of a pair is
3/51, because the first card dealt does not matter. Any card can pair up. The second card needs to be of the same rank as the first card. There are 3 cards left that satisfy this condition, but because one card is already dealt there are 51 cards left. Therefore the answer is
52/52 * 3/51 which is just
Probability of no pair is
48/51. Similarly the first card does not matter, but the second card needs to not pair, which are all remaining card minus three, which is
Probability of a suited hand is
12/51, since there are 13 cards of each suit in a deck of cards. The first card you get does not matter, but the second card needs to be of the same suit, which are 12 cards.
Probability of a non suited hand is
39/51. Same as the previous example, but now you need to essentially dodge 12 cards from the remaining 51 cards.
51 - 12 = 39.
Probability of a non suited non pair hand is
36/51, which indeed rougly equals
70.5%. Just like the previous example, but now you also need to not pair. Since these three cards are all different suits than the original cards we are not counting any cards twice, just substract three more.
Now we understand this logic we can tackle your problem. So we need to get dealt J9o. This means our first card has to be a jack or a nine, which are 8 cards total. Since there are 52 cards in a deck odds of this happening is
8/52. Now after that we need to hit the other card. There are 3 cards left that will get the job done, because suits need to be different. Odds of this happening therefore is
3/51. To calculate the odds of both these events occuring we need to multiply both numbers. We get
8/52 * 3/51 = 24/2652. We can simplify this to