Let Px and Py be the respective probabilities of Team X and Team Y winning in their own ballparks / Math please help

Question

Two baseball teams are tied for first place at the end of the season. According to tiebreaking rules, one of the teams will be designated as "Team X" and the other as "Team Y." Team X will then choose between two scenarios.

Scenario 1: The tied teams will play two games in Team X's ballpark. Team X must win BOTH games to be declared the champion. Otherwise, Team Y is the champion.

Scenario 2: The tied teams will play one game in Team Y's ballpark. The winner of this game will be declared the champion.

If P_y = 0.7, which inequality describes the values of P_x that would make scenario 1 the more favorable choice for Team X?

A. 0.35 < P_x </ 1

B. 0.49 < P_x </ 1

C. /0.3 <P_x </ 1

D. /0.7 < P_X</ 1

Answer 1

Px = Probability Tx wins at Fx (Tx Team X and Fx Field X)

Py = Prob Ty wins at Fy =0.7

Scenario 2: Py is in play, so Px = 0.3

Scenario 1: Probability of X champions is Px^2 because they have to win twice.

So for scenario 1 to be better than 0.3, Px must be greater than the square root of 0.3.

Since you have 4 answer options. Square the lower limit of each. You'll find that only one results with a value greater than 0.3. That is your answer.