A football team has a $75$% chance of victory per game, if their captain is in good form.
If the captain is in bad form, their chance of victory is $40$%.
In $70$% of the games, the captain is in good form.
I want to calculate the probability of the following:
- (a) the team wins a game
- (b) the captain is in good form, but the team doesn't win a game
Regarding (a), I think
$$ (0.75 \cdot 0.7) = 0.525 $$
Regarding (b), I think
$$ (0.7 \cdot 0.75) \cdot 0.4 = 0.21$$
Is this correct or wrong?
Best Answer
No this is wrong. Let "Captain in good form"=$C$
For part 1: $$P(Win)=1-P(Loss)$$ $$P(Loss)=P(Loss\cap C)+P(Loss\cap C')=P(Loss/C).P(C)+P(Loss/C').P(C')=0.25×0.7+0.6×0.3=0.355$$ Hence $P(Win)=0.645$
For part 2:
$$P(Loss/C).P(C)=P(Loss\cap C)$$ Hence $P(C\cap Loss)=0.25×0.7=0.175$