 Solution

A weakly dominant strategy must be a best response to either opponent action. For X this requiresThe first is stronger because $t>0$. ThusFor Y the comparisons are $t/3-P\ge 0$ and $(t-P)/2\ge 0$, givingBelow each upper endpoint the corresponding strategy is strictly better against both opponent actions; at the endpoint one comparison ties and the other is strict. Thus the answers also satisfy the convention that weak dominance requires a strict advantage somewhere. Since expected payoffs are linear in the opponent's mixing probability, checking both opponent pure strategies suffices for every opponent mixed strategy.