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

$The Reqd. Prob. = \frac{80}{99} \approx 0.8081 .$ $"The Reqd. Prob. "= 80/99~~0.8081.$

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