Question

. If a person visits his dentist, suppose the probability that he will have his teeth cleaned is 0.48, the
probability that he will have a cavity filled is 0.25, the probability that he will have a tooth extracted is
0.20, the probability that he will have teeth cleaned and a cavity filled is 0.09, the probability that he will
have his teeth cleaned and a tooth extracted is 0.12, the probability that he will have a cavity filled and a
tooth extracted is 0.07, and the probability that he will have his teeth cleaned, a cavity filled, and a tooth
extracted is 0.03. What is the probability that a person visiting his dentist will have at least one of these
things done to him?
Now, let C be the event that the person will have his teeth cleaned. Let F be the event of
getting cavity filled and E be the event of getting a tooth extracted.
Choose the correct option:
A) P(C⋂F)
I. 9/100
II. 7/10
III. 3/100
IV. 1/100
B) P(E)
I. 2/100
II. 1/100
III. 9/100
IV. None of these
C) P(C⋂E⋂F)
I. 3/100
II. 9/100
III. 1/100
IV. 12/100
D) P(E⋂F)
I. 7/100
II. 3/100
III. 9/100
IV. 11/100
E) P(C∪E∪F)
I. 68/100
II. 32/100
III. 16/100
IV. 2/100

Answer 1

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P(c) = 0.48 + 0.25 + 0.20 – 0.09 – 0.12 – 0.07 + 0.03 = 0.68

P(C) = 0.48, P (F) = 0.25, P (E) = .20, P (C ∩ F) = .09,

P (C ∩ E) = 0.12, P (E ∩ F) = 0.07 and P (C ∩ F ∩ Ε) = 0.03

Now, P ( C ∪ F ∪ E) = P (C) + P (F) + P (E) – P (C ∩ F) – P (C ∩ E) – P (F ∩ E) + P (C ∩ F ∩ E)

= 0.48 + 0.25 + 0.20 – 0.09 – 0.12 – 0.07 + 0.03 = 0.68

Answer 2

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Answer in attachment