Question

Suppose that Jamal can choose to get home from work by taxi or bus. When he chooses to get home by taxi, he arrives home after 7 p.m. 15 percent Of the time. When he chooses to get home by bus, he arrives home after 7 p.m. 8 percent of the time. Because the bus is cheaper, he uses the bus 70 percent of the time. What is the approximate probability that Jamal chose to get home from work by bus, given that he arrived home after 7 p.m.?

Answer 1

Answer:

0.55

Step-by-step explanation:

There are four things you need to keep in mind,and the question is too easy to solve , so following are those four things

1)Jamal travelling by Taxi before 7 PM

   It is given that Jamal uses the bus for 70 percent of the time so he will use the taxi for (100-70)%=30% of the time.

and also it is given that he arrives home by taxi after 7 PM , 15% of the time so he will arrive home by taxi before 7 PM ,(100-15)=85% of the time,

so by multiplication law of probability , the probability that Jamal will arrive by taxi before 7 PM =0.30*0.85=.255

2)Jamal travelling by Taxi after 7 PM

Similarly by considering the above situations as described in first part,

the probability that Jamal will arrive by taxi after 7 PM=0.30*0.15=.045

3) Jamal travelling by Bus before 7 PM

Now considering the case of the bus, it is given that Jamal arrives home after 7 PM 8 percent of the time. so for before 7 PM it will be 100-8= 92%

Similarly the probability that Jamal will arrive by bus before 7 PM=0.70*0.92=.644

4) Jamal travelling by Bus after 7 PM

Similarly on the basis of above explanation we can easily say that the probability that Jamal will arrive by bus after 7 PM=0.70*.08=.056

so, the probability that Jamal chose to get home from work by bus, given that he arrived home after 7 pm=$=\frac{Jamal travelling by bus after 7 PM}{(Jamal travelling by taxi after 7 PM +Jamal travelling by bus after 7 PM}$

$=\frac{0.056}{0.056+0.045}$

$=\frac{0.056}{0.101} \\=0.554$