Question

Carol lives in the east end of Montreal. To get to school for her morning classes Carol has the option of taking the bus, going underground with the metro, or driving her car downtown to get to Concordia. Carol prefers the metro, especially in the winter, and she will opt for it 35% of the time. The bus is perhaps a more convenient option since it stops near her house, so she will choose that option 55% of the time. The car is the most expensive option, but it is the most reliable, as demonstrated by the fact that she is only late for class 5% of the time when she drives. The bus is the least reliable of the options as it gets Carol to class on time 65% of the time, whereas this value increases by 10% with the metro.

c) What is the probability that Carol drove to school and is on time for class?
d) What is the probability that Carol opted for the bus given that she was late to class?
e) What is the probability that Carol is on-time for class given that she did not drive?

Answer 1

Answer on Question #60242 – Math – Statistics and Probability

Question

Carol lives in the east end of Montreal. To get to school for her morning classes Carol has the option of taking the bus, going underground with the metro, or driving her car downtown to get to Concordia. Carol prefers the metro, especially in the winter, and she will opt for it 35% of the time. The bus is perhaps a more convenient option since it stops near her house, so she will choose that option 55% of the time. The car is the most expensive option, but it is the most reliable, as demonstrated by the fact that she is only late for class 5% of the time when she drives. The bus is the least reliable of the options as it gets Carol to class on time 65% of the time, whereas this value increases by 10% with the metro.

a) What is the probability that Carol is late to class on a given morning?

b) What is the probability that Carol took the metro and is on time for class?

Solution

Denote by $H_1, H_2, H_3$ Carol decision to take bus, metro and car respectively. Then

$$P(H_1) = 0.55,$$ $$P(H_2) = 0.35,$$ $$P(H_3) = 0.1.$$

Also denote by $A$ the event that Carol will be late, so the condition entails

$$P(A|H_1) = 0.35,$$ $$P(A|H_2) = 0.25,$$ $$P(A|H_3) = 0.05.$$

a) Due to the formula of total probability

$$P(A) = P(A|H_1)P(H_1) + P(A|H_2)P(H_2) + P(A|H_3)P(H_3) = 0.55 \cdot 0.35 + 0.35 \cdot 0.25 + 0.1 \cdot 0.05 = 0.1925 + 0.0875 + 0.005 = 0.285 = 28.5\%.$$

b) $P(\widehat{A} \cap H_2) = P(\widehat{A}|H_2)P(H_2) = (1 - P(A|H_2))P(H_2) = (1 - 0.25)0.35 = 0.2625 = 26.25\%.$

Answer: a) 0.285; b) 0.2625.

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