Question

Suppose a basketball team had a season of games with the following characteristics:

• Of all the games, 60% were at-home games. Denote this by H (the remaining were away games).

• Of all the games, 25% were wins. Denote this by W (the remaining were losses).

• Of all the games, 20% were at-home wins.

Of the at-home games, what proportion of games were wins? (Note: Some answers are rounded to two decimal places.)

A 0.12

B 0.15

C 0.20

D 0.33

E 0.42

Answer 1

Given, P(at-home games) = 0.60 ; P(wins) = 0.25 ; P(at-home games and wins) = 0.20

P(wins | at-home games) = P(at-home games and wins) / P(at-home games)            ...........(conditional probability)

                                         = 0.20 / 0.60

                                         = 0.33

Therefore, of the at-home games, the proportion of games which were wins is 0.33