Question

ENGR 315, Summer 2019 Name 3. Supp0se that Tom, an ENGR 315 commuter student, can decide to get to the Widener campus by one of three modes of transportation, i.e. car (C), bus (B), or train(T). If he decides to go by car, there will be a 50% chance that he will be late. If he goes by bus, the probability of being late is 20%. If he travels by train, the probability of being late is only 1 % ( but is more expensive). A. Draw a Venn diagram depicting this situation. (5 pts.) B. Suppose that Tom is late one day, and his teacher wishes to estimate the probability that he drove to campus that day by car. The teacher does not know which mode of transportation Tom usually uses, so he gives a prior probability of 1/3 to each of the three known possibilities. What is the teacher's estimate of the probability that Tom drove to school? (15 pts.) C. Suppose that one of Tom's classmates knows that he almost alwavs takes the train to school, never takes the bus, but sometimes (10%% ) takes his car. What is the classmate's probability that Tom drove to school that day, given that he was late? (5 pts.)

Answer 1

Let E₁, E₂ and E​​​​₃, be the events of going by car, bus and train respectively and L be the event of getting late.

A.

Venn diagram:

E3 E2 EL Bus Train Car La te LMES (LnE, LAEZ

B.

We know that,

P(E₁) = P(E₂) =P(E₃) = 1/3

P(L|E₁) = 0.5

P(L|E₂) = 0.2

P(L|E₃) = 0.01

P(L) = P(L|E₁)*P(E₁) + P(L|E₂)*P(E₂) + P(L|E₃)*P(E₃)

= 0.5*1/3 + 0.2*1/3 + 0.01*1/3

= 0.2367

P(L|E₁) = 0.5

P(L$\cap$E₁)/P(E₁) = 0.5

P(L$\cap$E₁) = 0.5/3

= 0.1667

P(E₁|L) = P(L$\cap$E₁)/P(L)

= 0.1667/0.2367

= 0.7043

C.

We know that,

P(E₁) = 0.1

P(E₂) = 0

P(E₃) = 0.9

P(L|E₁) = 0.5

P(L|E₂) = 0.2

P(L|E₃) = 0.01

P(L) = P(L|E₁)*P(E₁) + P(L|E₂)*P(E₂) + P(L|E₃)*P(E₃)

= 0.5*0.1 + 0.2*0 + 0.01*0.9

= 0.059

P(L|E₁) = 0.5

P(L$\cap$E₁)/P(E₁) = 0.5

P(L$\cap$E₁) = 0.5*0.1

= 0.05

P(E₁|L) = P(L$\cap$E₁)/P(L)

= 0.05/0.059

= 0.8475

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