Question on Independent Events

Question

I have this example from the All of Statistics book:

I'm looking for an explanation for P(A2). Why is it equal to (2/3)(3/4)(1/3)?

Thanks

Answer 1

Symbolically,

\begin{align} P(A_2) & = P(\text{$P_1$ succeeds before $P_2$, on trial $2$}) \\ & = P(\text{$P_1$ fails on trial $1$ and succeeds on trial $2$}) \times P(\text{$P_2$ fails on trial $1$}) \\ & = P(\text{$P_1$ fails on trial $1$}) \times P(\text{$P_1$ succeeds on trial $2$}) \times P(\text{$P_2$ fails on trial $1$}) \\ & = \frac23 \times \frac13 \times \frac34 = \frac16 \end{align}

Keep in mind that the players take turns, so Player $1$ takes their second turn before Player $2$ does.