Please use Union and intersection not a formula. In your town, it is raining one

Question

Please use Union and intersection not a formula. In your town, it is raining one third of the days. Given that it is raining, there will be heavy traffic with a probability of 0.50, and given that it is not raining, there will be heavy traffic with a probability of 0.25. If it is raining and there is heavy traffic, you will arrive late to class with a probability of 0.5. On the other hand, the probability of being late is 0.125 if it is not raining and there is no heavy traffic. In other situations (rain with no traffic and traffic without rain), the probability of being late to class is 0.25. On a random day, determine (a) the probability that it is not raining and there is heavy traffic, but you are not late to class, (b) the probability that you are late to class, and (c) given that you were late to class, it was raining?

Explanation / Answer

Let R denote raining and Rc denote its complement

Let H denote heavy traffic and Hc its complement

L denote late arrival and Lc its complement.

We are given

P(R)=1/3

P(H/R)=0.50

P(H/Rc)=0.25

P(L/RH)=0.5

P(L/RcHc)=0.125

P(L/(RHc and RcH)=0.25 so that P(L/RHc)+P(L/RcH)=0.25 as RHc and RcH are disjoint.

We have to find P(RcHLc), P(L) and P(R/L)

Part-a

P(RcHLc)=P(Lc/RcH)*P(RcH)

= [1- P(L/RcH)]*P(RcH)

=[1-0.25+ P(L/RHc)] *P(RcH)

=0.75*P(RcH) + P(L/RHc)*P(RcH)

Now, P(RcH)=P(Rc)*P(H/Rc) =(1-1/3)*0.25 = 0.5/3

P(RHc)=P(R)*P(Hc/R)=(1/3)*(1-0.5)=0.5/3

P(L/RHc)=P(LRHc)/P(RHc)=P(Hc/LR)/P(RHc) = [1-P(H/LR)] P(RHc)=[1-P(LHR)/P(LR)]P(RHc)

=[1-P(H)]*P(Rhc)

=(1-1/3)*0.5/3

=1/9

So, P(RcHLc)=0/75*0.5/3+(1/9)*0.5/3=0.0185

Part-b

P(L)=P(L/HR)*P(HR)

Now, P(HR)=P(H/R)*P(R)=0.50*(1/3)

So, P(L)=0.5*0.50*(1/3) =0.0833

Part-c

P(R/L) =P(RL)/P(L)

P(RL)=P(L/R)=P(L/RH)+P(L/RHc)=0.5+0.25-P(L/RcH)

Now, P(L/RcH)= P(L/RH)- P(L)=0.5-0.0833

So, P(RL)=0.5+0.25-0.5+0.0833 =0.3333

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