Event: EH = doing excellent homework
if GB, then P(EH) = 0.8.
if not GB, then P(EH) = 0.5.
A) For a random week, calculate the probability that :
(i) he will be given a special treat;
He is given a special treat whenever whenever his report indicates "good behaviour" AND "Excellent homework".
i.e. both events must happen.
i.e. GB and EH
That probability = P(GB) * ( P(EH), if GB happens) = 0.6 * 0.8 = 0.48
(ii) his homework will be excellent but he will be NOT be given a special treat;
Aha! That means that his behaviour is not good, otherwise, he would have got a treat.
So, this happens whenever whenever his report indicates NOT "good behaviour" AND "Excellent homework".
i.e. both events must happen.
i.e. not GB and EH
That probability = P(Not GB) * ( P(EH), if not GB happens)
= ( 1- 0.6) * 0.5 = 0.2
(iii) his homework will be excellent.
Two situations:
x) his homework is excellent AND his report indicates "good behaviour".
y) his homework is excellent AND his report indicates NOT "good behaviour".
We just calculated these two scenarios. Either one can happen, so it is
0.48 + 0.2 = 0.68
b) Given that his homework is excellent, calculate the conditional probability that he is NOT given a special treat.
Aha! That means that his behaviour is not good, otherwise, he would have got a treat.
So, P( EH, given GB) = 0.48
So, P( EH, given NOT GB) = 0.2
So, if we know EH happened, then probability of NOT GB =
P( EH, given NOT GB) /
{ P( EH, given GB) + P( EH, given NOT GB) }
= 0.2 / { 0.48 + 0.2 } = 0.2/0.68
