Question

A rep hockey team has two goalies. Goalie A has a save percentage of 0.92 and Goalie B has a save percentage of 0.89. If goalie A plays 75 percent of the games and goalie B plays 25 percent of the games then: a) Determine the probability that a save has been made by either goalie A or goalie B. b) If a save has been made, what is the probability that is was goalie A? c) If a goal was let in, what is the probability it was goalie B?

Answer 1

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.