Question

"Based on previous baseball games, a sports statistician has calculated that there is a 38% chance of Team A beating Team B. Team A and Team B are scheduled to play each other 10 times this season. If each game is played independently of the others, what is the probability that Team A will beat Team B for the first time at the 10th game?

A) 0.38
B) 0.62
C) (0.38)10
D) (0.38)9 (0.62)
E) (0.62)9 (0.38)

Answer 1

The probability that Team A will beat Team B for the first time on the 10th game is found by calculating the probability of losing 9 times and then winning once, which is (0.62)⁹ × (0.38), answer E.

Step-by-step explanation:

The probability that Team A will beat Team B for the first time at the 10th game requires Team A to lose the first 9 games and then win the 10th game. Since each game is independent, you can calculate this probability by multiplying the probability of losing 9 times in a row with the probability of winning on the 10th try. This calculation is performed as follows: (0.62)⁹ × (0.38), where 0.62 is the probability of Team A losing a game (100% - 38% win probability).

The correct answer is therefore E) (0.62)⁹ × (0.38).