Answer:
a) 0.9125
b) 0.756
c) 0.314
Step-by-step explanation:
The save percentage of Goalie A = 0.92
The save percentage of Goalie B = 0.89
The percentage of the games Goalie A plays = 75 percent
The percentage of the games Goalie B plays = 25 percent
a) The probability that a save has been made by either Goalie A or Goalie B
Therefore we get;
The probability that Goalie A made a save, P(A) = 0.92 × 0.75 = 0.69
The probability that Goalie B made a save, P(B) = 0.89 × 0.25 = 0.2225
For mutually exclusive events, we have;
P(A) or P(B) = P(A) + P(B)
∴ P(A) or P(B) = 0.69 + 0.2225 = 0.9125
The probability that a save has been made by either Goalie A or Goalie B, P(A) or P(B) = 0.9125
b) Where there are 10,000 attempts, made evenly in all games, we have;
The number of attempts made with Goalie A playing = 10,000 × 0.75 = 7,500
The number of saves Goalie A makes = 0.92 × 7,500 = 6,900
The number of saves for 10,000 attempt = (P(A) or P(B)) × 10,000 = 0.9125 × 10,000
The number of saves for 10,000 attempt = 9,125
∴ The probability that Goalie A saves = 6,900/9,125 ≈ 0.756
c) Out of the 2,500 attempts at Goalie B, the number of goals let in is given as follows;
P(Goal B) = 2,500 - 2,500 × 0.89 = 275
The number of goals let in by Goalie A, we have;
P(Goal A) = 7,500 - 7,500 × 0.92 = 600
Therefore, the total number of goals let in = 300 + 275 = 875
The probability that the goals was let in by Goalie B = 275/875 = 11/35 ≈ 0.314.
