I'll explain it a bit clearer.
Each basher has a 15% chance to bash (assuming melee).
p=0.15

If you have 2 bashers, each of them has a 0.15 chance to bash, independant of each other. You have 4 possible outcomes.

First basher bash, second doesn't.
First basher bash, second bash as well
First basher doesn't bash, second bashes.
First basher doesn't bash, second doesn't as well.

Now let's name them equation 1 to 4, according from top to bottom. The chances of each event happening...

1)0.15*0.85
2)0.15*0.15
3)0.85*0.15
4)0.85*0.85

Now we want to calculate what is the chance of bashing. The criteria is and/or, as long as one bash activates (doesn't matter if both activates at the same time as well), it is considered a 'bash'.

So therefore, we take 1 + 2 + 3, in other words

0.15*0.85 + 0.15*0.15 + 0.85*0.15=0.2775 i.e 27.75% chance (alternatively you can just take 1 - * figure obtained from 4, 0.85*0.85 *)

I'll explain it a bit clearer.
Each basher has a 15% chance to bash (assuming melee).
p=0.15

If you have 2 bashers, each of them has a 0.15 chance to bash, independant of each other. You have 4 possible outcomes.

First basher bash, second doesn't.
First basher bash, second bash as well
First basher doesn't bash, second bashes.
First basher doesn't bash, second doesn't as well.

Now let's name them equation 1 to 4, according from top to bottom. The chances of each event happening...

1)0.15*0.85
2)0.15*0.15
3)0.85*0.15
4)0.85*0.85

Now we want to calculate what is the chance of bashing. The criteria is and/or, as long as one bash activates (doesn't matter if both activates at the same time as well), it is considered a 'bash'.

So therefore, we take 1 + 2 + 3, in other words

0.15*0.85 + 0.15*0.15 + 0.85*0.15=0.2775 i.e 27.75% chance (alternatively you can just take 1 - * figure obtained from 4, 0.85*0.85 *)