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@kropot72 @jim_thompson5910
@zepdrix
what is the probability they played at home?
36/100?
or 0.36, yes
then 56/81?
wait, that's not correct
oh ok
they won 36 games at home, out of 100 so P(winning at home) = 0.36 but I wanted the probability of playing at home
well the good news is that 0.36 is for `the likelyhood they played at home, given they won` since the `given they won` means we only focus on the 100 games they won
The team played 156 games They won 100 of them They lost 156-100 = 56 games --------------------------- The team played 81 away games So the team played 156-81 = 75 home games --------------------------- Home games The team won 36 home games The team lost 75 - 36 = 39 home games --------------------------- Away games: The team won 100 games. The team won 36 home games, so they must have won the remaining 100-36 = 64 games on the road The other 81-64 = 17 games played on the road were losses Do you agree with everything so far?
yes wow thank you so much
ok we'll use this info to find the probabilities
`the likelihood the team won given they played at home` things you'll need to find * how many home games were played total? * how many wins were at home?
36/156?
how many home games were played total?
So the team played 156-81 = 75 home games
yes so we have 36 home wins out of 75 home games the probability they won, given they're at home, is 36/75
the "given at home" part means you only focus on the home games
that would be for the first one correct?
it might help to make a table |dw:1460681570962:dw|
|dw:1460681623551:dw|
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