A dental insurance policy covers three procedures: orthodontics, fillings, and extractions. During the life of the policy, the probability that the policyholder needs
orthodontic work is $1/2$;
orthodontic work or a filling is $2/3$;
orthodontic work or an extraction is $3/4$,
a filling and an extraction is $1/8$.
The need for orthodontic work is independent of the need for a filling and is independent of the need for an extraction. Calculate the probability that the policyholder will need a filling or an extraction during the life of the policy.
My attempt:Let $O, F, E$ denote the events that a policyholder would need orthodontic work, a filling, or an extraction respectively. If $A$ is an event, let $A'$ denote the complement of $A$.
Now, $P(O\cup E) = 1-P(O'\cap E') = 1-\left[P(O')\cdot P(E')\right] = 1-\left[\dfrac{ P(E')}{2}\right] \implies \dfrac{3}{4} = 1-\left[\dfrac{ P(E')}{2}\right] \implies P(E') = \dfrac{1}{2}$
Similarly, $P(O\cup F) = 1-P(O'\cap F') = 1-\left[P(O')\cdot P(F')\right] \implies \dfrac{2}{3} = 1-\left[\dfrac{ P(F')}{2}\right]\implies P(F') = \dfrac{2}{3}$
So, $P(F \cup E) = 1-P(F' \cap E') = 1-\left[\dfrac{2}{3} \cdot \dfrac{1}{2}\right] = \dfrac{2}{3}$, which is, unfortunately, not the correct response. Can someone please review my solution and let me know where I went wrong? Thanks!
$\endgroup$1 Answer
$\begingroup$You correctly got $P(F')=2/3$, i.e. $P(F)=1/3$ and $P(E')=1/2$ i.e. $P(E)=1/2$.
Your mistake was then that you assumed the need for filling and the need for extraction were independent. That is not mentioned in the problem. Had those events been independent, you would've had $P(F\cap E)=P(F)P(E)=1/6$, but it is not that - it is $1/8$.
Instead, you should've just calculated $P(F\cup E)=P(F)+P(E)-P(F\cap E)=1/2+1/3-1/8=17/24$.
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