Respuesta :
Answer:
B-91%
Step-by-step explanation:
Bayes Theorem:
Two events, A and B.
[tex]P(B|A) = \frac{P(B)*P(A|B)}{P(A)}[/tex]
In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.
In this question:
Event A: arriving home after 7 p.m.
Event B: getting home by bus.
When he chooses to get home by bus, he arrives home after 7 p.m. 25 percent of the time.
This means that [tex]P(A|B) = 0.25[/tex]
Because the bus is cheaper, he uses the bus 70 percent of the time.
This means that [tex]P(B) = 0.7[/tex]
Probability of getting home after 7 p.m.
70% of the time he uses bus, and by bus, he arrives arrives home after 7 p.m. 25 percent of the time.
100 - 70 = 30% of the time he uses the car, and by car, he arrives home after 7 p.m. 6 percent of the time.
So
[tex]P(A) = 0.7*0.25 + 0.3*0.06 = 0.193[/tex]
What is the approximate probability that Matthew chose to get home from work by bus, given that he arrived home after 7 p.m.?
[tex]P(B|A) = \frac{0.7*0.25}{0.193} = 0.9067[/tex]
Rouding up, 91%.
So the correct answer is:
B-91%
