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

Hello Sir/ Mam

YOUR REQUIRED ANSWER IS 18.6%

Variables defined in the question:

H - Home - Games

W - Winning Games

Now, given:

P(H0.62 PW)0.22

P(W\mid H)=0.30

Required : Of all the games, percentage of games that were at home wins.

Hence, required :

P(W\cap H)

Now, we know that:

P(W\mid H)=\frac{P(W\cap H)}{P(H)}

0.3=\frac{P(W\cap H)}{0.62}

Hence,

P(W\cap H)=0.3*0.62=0.186=18.6\%

I hope this solves your doubt.

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