The probability of having a toothache because of a cavity

Yandel Hunter

Yandel Hunter

Answered question

2022-02-11

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, whats

Answer & Explanation

glavninamvyw

glavninamvyw

Beginner2022-02-12Added 11 answers

Step 1
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(TC)=0.8=45,
P(C)=0.05=120,
and
P(TC)=0.01=1100,
and we want, P(CT)
Recall that, P(AB)=P(AB)P(B)
P(TC)=45P(TC)P(C)=45
P(TC)=(120)(45)P(TC)=125 (1)
Similarly, P(TC)=P(TC)P(C)=P(TC)1P(C)
1100=P(TC)1120P(TC)=(1100)(1920),i.e.,
P(TC)=192000 (2)
Now, (TC)(TC)=T(CC)Tϕ=ϕ(),&
(TC)(TC)=T(CC)=TU=T()
Thus, from ()&(), we see that, (TC)&(TC)
are mutually exclusive events and T is their Union Event.
Therefore, by (1)&(2), we get,
P(T)=P(TC)+P(TC)=125+192000=992000

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