Question

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?

Answer 1

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%