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?

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Answer 1 · 2014-12-25T08:46:16

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: $P(H)=0.6$
Loosing home $P(L)=1-0.7=0.3$
Multiply because it is $H andL$
$P(HL)= 0.6*0.3=0.18$

(2) Playing out: $P(O)=1-0.6=0.4$
Loosing out $P(L)=1-0.35=0.65$
Multiply because it is $O andL$
$P(OL)=0.4*0.65=0.26$

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

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