Resolved: "The probability of having a toothache because of..."

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

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.
$P (\frac{T}{C}) = 0.8 = \frac{4}{5}$ ,
$P (C) = 0.05 = \frac{1}{20}$ ,
and
$P (\frac{T}{C^{'}}) = 0.01 = \frac{1}{100}$ ,
and we want, $P (\frac{C}{T})$
Recall that, $P (\frac{A}{B}) = \frac{P (A \cap B)}{P (B)}$
$∴ P (\frac{T}{C}) = \frac{4}{5} \Rightarrow \frac{P (T \cap C)}{P (C)} = \frac{4}{5}$
$\Rightarrow P (T \cap C) = (\frac{1}{20}) (\frac{4}{5}) \Rightarrow P (T \cap C) = \frac{1}{25}$ (1)
Similarly, $P (\frac{T}{C^{'}}) = \frac{P (T \cap C^{'})}{P (C^{'})} = \frac{P (T \cap C^{'})}{1 - P (C)}$
$\Rightarrow \frac{1}{100} = \frac{P (T \cap C^{'})}{1 - \frac{1}{20}} \Rightarrow P (T \cap C^{'}) = (\frac{1}{100}) (\frac{19}{20}), i . e .,$
$P (T \cap C^{'}) = \frac{19}{2000}$ (2)
Now, $(T \cap C) \cap (T \cap C^{'}) = T \cap (C \cap C^{'}) T \cap ϕ = ϕ \dots (\cdot), &$
$(T \cap C) \cup (T \cap C^{'}) = T \cap (C \cup C^{'}) = T \cap U = T \dots (⋆)$
Thus, from $(\cdot) & (⋆)$ , we see that, $(T \cap C) & (T \cap C^{'})$
are mutually exclusive events and T is their Union Event.
Therefore, by $(1) & (2)$ , we get,
$P (T) = P (T \cap C) + P (T \cap C^{'}) = \frac{1}{25} + \frac{19}{2000} = \frac{99}{2000}$

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