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, we are interested in finding what proportion were wins. Which of the following probabilities do you need to find in order to determine the proportion of at-home games that were wins?
A. P(H)
B. P(W)
C. P(H and W)
D. P(H | W)
E. P(W | H)

Answer 1

Answer:

E. P(W | H).

Step-by-step explanation:

We are given that 60% are at-home games. So,

P(H)=0.60.

We are given that 25% are winning games. So,

P(W)=0.25.

We are given that 20% are at-home winning games. So,

P(H and W)=0.20.

We are interested in finding what proportion of at-home games wins. So, we want are given that basketball team had a home game and we want to find win games. So, we are interested in finding probability of win given that a game is at-home game. Thus, we want to determine P(W|H).

Further, P(W|H) can be determine with the given information as

P(W|H)=P( W and H)/P(H)

P(W|H)=0.2/0.6

P(W|H)=0.333.