# A basketball team plays 60% of its games at home. The home court advantage is obvious, because the team wins 70% of its home games, but when they play away, they win only 35% of their games. What is the (conditional) probability of losing, GIVEN THAT the game was played away?

Dec 25, 2014

We will have to consider two distinct cases: home (H) and out (O) within these cases we have win (W) and loose (L)

We note probabilities by P and convert % to fractions
(divide by 100)

(1) Playing home: $P \left(H\right) = 0.6$
Loosing home $P \left(L\right) = 1 - 0.7 = 0.3$
Multiply because it is $H \mathmr{and} L$
$P \left(H L\right) = 0.6 \cdot 0.3 = 0.18$

(2) Playing out: $P \left(O\right) = 1 - 0.6 = 0.4$
Loosing out $P \left(L\right) = 1 - 0.35 = 0.65$
Multiply because it is $O \mathmr{and} L$
$P \left(O L\right) = 0.4 \cdot 0.65 = 0.26$

Since cases (1) and (2) are of the "either...or" type you may add the probabilities:
$P \left(L\right) = P \left(H L\right) + P \left(O L\right) = 0.18 + 0.26 = 0.44$

Conclusion:
The (conditional) probability of loosing is 44%