The probability of having a toothache because of a cavity is 0.8. Suppose the probability of having a cavity is 0.05. Assuming the probability of a toothache given you have no cavity is 0.01, what's 'the probability of a cavity if you have a toothache?

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Answer 1 · 2017-01-18T11:15:13

#"The Reqd. Prob. "= 80/99~~0.8081.#

Let #T# be the event that you have Toothache , and, #C# be the

event that you have Cavity . Then, #C'# denotes the event that

you have No Cavity.

In the Usual Notation of Conditional Probability , we have,

#P(T/C)=0.8=4/5, P(C)=0.05=1/20, and, P(T/(C'))=0.01=1/100,# and, we want, #P(C/T).#

Recall that, #P(A/B)=(P(AnnB))/(P(B))#

#:. P(T/C)=4/5 rArr (P(TnnC))/(P(C))=4/5#

#rArr P(TnnC)=(1/20)(4/5) rArr P(TnnC)=1/25..........(1)#

#"Similarly, "P(T/(C'))={P(TnnC')}/{P(C')}={P(TnnC')}/{1-P(C)}#

#rArr 1/100={P(TnnC')}/(1-1/20) rArr P(TnnC')=(1/100)(19/20), i.e.,#

#P(TnnC')=19/2000..........................(2)#

Now, #(TnnC)nn(TnnC')=Tnn(CnnC')Tnnphi=phi...(ast),# &,

#(TnnC)uu(TnnC')=Tnn(CuuC')=TnnU=T.........(star)#

Thus, from #(ast) & (star)#, we see that, #(TnnC) & (TnnC')#

are mutually exclusive events and #T# is their Union Event.

Therefore, by #(1) & (2),# we get,

#P(T)=P(TnnC)+P(TnnC')=1/25+19/2000=99/2000.#

Finally, the Reqd. Prob. #=P(C/T)={P(CnnT)}/{P(T)}#

#=(1/25)/(99/2000)=80/99~~0.8081 (4dp). #

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