Question #a7582

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Answer 1 · 2016-11-12T12:32:35

The overall probability of being late $p(L)=0.17$
The probability of not late and commuting by car $=24/83$

(1) Let's calculate the probability of johnny being late

train = T
car = C
bike = B
late =L
not late = $barL$

$p(TnnL)=0.2*0.3=0.06$

$p(CnnL)=0.3*0.2=0.06$

$p(BnnL)=0.5*0.1=0.05$

$p(L)=p(TnnL)+p(CnnL)+p(BnnL)=0.06+0.06+0.05=0.17$

The overall probability of being late $p(L)=0.17$

(2)Probability of Johnny not being late $p(barL)=1-0.17=0.83$

$(p(barL nn C))=0.3*0.8=0.24$

We need to calculate $p_(barL)(C)=(p(barL nn C))/(p(barL))=(0.8*0.3)/0.83=0.24/0.83$