Three-person jury with a coin-flipping member

Panel A has three members deciding by majority vote: two of them are each independently right with probability $p$, while the third decides every case by tossing a fair coin. Panel B is one person who is right with probability $p$. Which panel is more likely to reach the correct verdict?

Neither — both are correct with probability $p$. Panel A is right when the two serious members both are right ($p^2$), or when they split ($2p(1-p)$) and the coin sides with the right one (half the time). That totals $p^2+p(1-p)=p$.