PLEASE ANSWER ALL QUESTIONS!!! I. A car insurance company insures four types of

Question

PLEASE ANSWER ALL QUESTIONS!!!

I. A car insurance company insures four types of drivers: . Good drivers who have a 2% chance of getting in an accident each year Mediocre drivers have a 5% chance of getting in an accident in each year Atrocious drivers who have a 15% chance of getting in an accident in each year Ludicrously bad drivers who have a 60% chance of getting in an accident each year 60% of drivers are good, 25% are mediocre, 13% are atrocious, and 2% are ludicrously bad. A driver they insure gets in an accident, what is the likelihood he is a good mediocre, atrocious, or ludicrously bad driver. 2. Identifv the independent events: P(A)-5 P(B)-.5 P(AUB)-.75 3. I roll two dice, let A be the event the sum of the dice is a perfect square (i.e. 4, or 9) and B be the event the sum of the dice is an even number. Are A and B independent? Every day your friend commutes to school on the subway at 9 on time, she will stop for a $3 coffee on the way to class. If the subway is delayed she skips the coffee and goes straight to class. The probability that the subway is delayed is 40%. Find the probabilities that she spends. 0. 3, 6, 9, 12, and 15 dollars on coffee over the course of a five day week 4. AM. If the subway i:s

Explanation / Answer

1.
A: Random Driver is good
B: Random Driver is Mediocre
C: Random Driver is Atrocious
D: Random Driver is ludicrously bad
P(A) = 0.60, P(B) = 0.25, P(C) = 0.13, P(D) = 0.02
X: Driver gets into an accident.
Chances of a good driver getting into an accident: P(X | A)
Chances of a mediocre driver getting into an accident: P(X | B)
Chances of a atocious driver getting into an accident: P(X | C)
Chances of a ludicrous driver getting into an accident: P(X | D)
P(X | A) = 0.02, P(X | B) = 0.05, P(X | C) = 0.15, P(X | D) = 0.60
A random driver gets in an accident:
Probability that he is a good driver: P(A | X)
Using conditional probability: P(A | X) = P(A X) / P(X)
P(A | X) = P(A).P(X | A) / P(X)
Using total law, first calculate P(X) = P(A).P(X | A) + P(B).P(X | B) + P(C).P(X | C) + P(D).P(X | D)
P(X) = (0.60 x 0.02) + (0.25 x 0.05) + (0.13 x 0.15) + (0.02 x 0.60) = 0.056
P(A | X) = (0.60 x 0.02) / 0.056 = 0.214

Probability that he is a mediocre driver: P(B | X) = P(B).P(X | B) / P(X) = (0.25 x 0.05) / 0.056 = 0.223

Probability that he is a good driver: P(C | X) = P(C).P(X | C) / P(X) = (0.13 x 0.15) / 0.056 = 0.348

Probability that he is a good driver: P(D | X) = P(D).P(X | D) / P(X) = (0.02 x 0.60) / 0.056 = 0.214

2.
Two events A and B are independent if P(A B) = P(A).P(B)

P(A) = 0.5 P(B) = 0.5

P(A U B) = P(A) + P(B) - P(A B)

P(A B) = P(A) + P(B) - P(A U B) = 0.5 + 0.5 - 0.75 = 0.25

P(A).P(B) = 0.25 = P(A B)

Hence A and B are independent.

P(C D) = 0.16

P(C).P(D) = 0.2 x 0.8 = 0.16

Hence C and D are independent events.

P(Ec Fc) = 0.3
P(E) = 0.4 => P(Ec) = 0.6
P(F) = 0.5 => P(Fc) = 0.5
P(Ec).P(Fc) = 0.3 = P(Ec Fc)
Hence E and F are independent.

P(G | H) = 0.65 , P(G | Hc) = 0.35 Clearly G depends on the occurence of event H, hence G and H are not independent.

P(I | J) = 0.7 , P(Ic | Jc) = 0.3 => P(I | Jc) = 1 - P(Ic | Jc) = 1-0.3 = 0.7
Since P(I | J) = P(I | Jc) , event I does not depend on the occurence of event J. Hence I and J are independent events.

3.
P(A) = 2/36 (only possibilities are 4 and 9)
P(B) = 18/36
P(A B) = 1/36 (case when sum of dice is 4)
P(A).P(B) = 2/36 x 18/36 = 1/36 = P(A B)
Hence A and b are independent events.

4.
Let X denote the event that subway is delayed.

P(X) = 0.40

Probability that she spends 0 dollars on coffee over the course of a 5 day week = Probability that the subway is delayed on each of the 5 days = P(X)5 = 0.405 = 0.0102

Probability that she spends 3 dollars on coffee over the course of a 5 day week = Probability that the subway is delayed on any 4 of the 5 days = (5C4) x P(X)4 x P(Xc) = 0.0768

Probability that she spends 6 dollars on coffee over the course of a 5 day week = Probability that the subway is delayed on any 3 of the 5 days = (5C3) x P(X)3 x P(Xc)2 = 0.2304

Probability that she spends 9 dollars on coffee over the course of a 5 day week = Probability that the subway is delayed on any 2 of the 5 days = (5C2) x P(X)2 x P(Xc)3 = 0.3456

Probability that she spends 12 dollars on coffee over the course of a 5 day week = Probability that the subway is delayed on any 1 of the 5 days = (5C1) x P(X)1 x P(Xc)4 = 0.2592

Probability that she spends 15 dollars on coffee over the course of a 5 day week = Probability that the subway is not delayed on any of the 5 days = P(Xc)5 = 0.0778

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