Suppose that Kevin can choose to get home from work by car or bus. When he chooses to get home by car, he arrives home after 7 p. M.4 percent of the time. When he chooses to get home by bus, he arrives home after 7 p. M. 15 percent of the time. Because the bus is cheaper, he uses the bus 58 percent of the time. What is the approximate probability that Kevin chose to get home from work by bus, given that he arrived home after 7 p. M.?

The approximate probability that Kevin chose to get home from work by bus, given that he arrived home after 7 pm = 0.838

Step-by-step explanation:

Let the probability that Kevin arrives home after 7 pm be P(L)

Probability that Kevin uses the bus = P(B)

Probability that Kevin uses the car = P(C)

Probability of arriving home after 7 pm if the car was taken = P(L|C) = 4% = 0.04

Probability of arriving home after 7 pm if the bus was taken = P(L|B) = 15% = 0.15

The bus is cheaper, So, he uses the bus 58% of the time.

P(B) = 58% = 0.58

P(C) = P(B') = 1 - P(B) = 1 - 0.58 = 0.42

The approximate probability that Kevin chose to get home from work by bus, given that he arrived home after 7 pm = P(B|L)

The conditional probability P(A|B) is given mathematically as

P(A|B) = P(A n B) ÷ P(B)

Hence, the required probability, P(B|L) is given as

P(B|L) = P(B n L) ÷ P(L)

But we do not have any of P(B n L) and P(L)

Although, we can obtain these probabilities from the already given probabilities

P(L|C) = 0.04

P(L|B) = 0.15

P(B) = 0.58

P(C) = 0.42

P(L|C) = P(L n C) ÷ P(C)

P(L n C) = P(L|C) × P(C) = 0.04 × 0.42 = 0.0168

P(L|B) = P(L n B) ÷ P(B)

P(L n B) = P(L|B) × P(B) = 0.15 × 0.58 = 0.087

P(L) = P(L n C) + P(L n B) = 0.0168 + 0.087 = 0.1038 (Since the bus and the car are the two only options)

The approximate probability that Kevin chose to get home from work by bus, given that he arrived home after 7 pm

= P(B|L) = P(B n L) ÷ P(L)

P(B n L) = P(L n B) = 0.087

P(L) = 0.1038

P(B|L) = (0.087/0.1038) = 0.838150289 = 0.838

Hope this Helps!!!

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