I'm not quite sure I'm getting this right, as you don't go into much detail in your question:
You show two truth tables for "p implies q", or "p => q". (your variable names are p2 and p3, and p3 and p1 respectively). In a truth table you have a list of all possible combinations of values for the variables. With two boolean values you have four possible combinations, TT/TF/FT/FF (T is represented by 1, and F by 0 in your tables).
The rule for '=>' is that it is true unless the precedent is false and the output is true. For example, if we use it rains for p, and the road is wet for q, we get the following table:
- 0 0 IF it DOES NOT rain THEN the road is NOT wet (true)
- 0 1 IF it DOES NOT rain THEN the road is wet (true)
- 1 0 IF it rains THEN the road is NOT wet (false)
- 1 1 IF it rains THEN the road is wet (true)
This mirrors your truth tables above. Even though it might seem counter intuitive, only line 3 is false (0). Line 1 and 4 are obvious, and line 2 states that even if it does not rain, the road could still be wet (by someone using a hose pipe to make it wet).
So in the implication, 3 of the possible 4 combinations result in a true outcome. Which is why you get "Yes: 3".