Afriend who works in a big city owns two cars, one small and one large. one-quarter of the time he drives the small car to work, and three-quarters of the time he takes the large car. if he takes the small car, he usually has little trouble parking and so is at work on time with probability 0.9. if he takes the large car, he is on time to work with probability 0.6. given that he was at work on time on a particular morning, what is the probability that he drove the large car?

P (A║B)  = 0.54

Step-by-step explanation: To solve this problem we will use Bayes Theorem

express as:

P (A║B)  =   P (A) *  P (B║A)  /  P(B)

A friend has two car                                   smaller           and            large

Probability of using a car                          1/4 = 0.25                   3/4  =  0,75

probabilities arriving on time Event (A)           0.9                               0.6

Event B use of large car

We are going to examine the probabilities of arriving on time

If he uses the small car   his probabilities are  = 0,25*0.9 =  0.225

If he uses the large  car   his probabilities are  = 0.75 * 0.6 = 0.45

Total probability of arriving on time is =  0.225  +  0.45  = 0,675

Probability of arriving on time given that he drove a lage car  = 0.6

Probability of using a large car is 0.75

Then by subtitution we get

P (A║B)  =  ( 0.675 ) * 0.6 / 0.75

P (A║B)  = 0.54

two data sets with the same median will not have the same varaibility

step-by-step explanation:

answer: the answer is 30 miles per hour.

ft/s -> mphmph = ft/s 1.466667mph = 44 divided by 1.466667mph = 30.00000

- r3ktforgood ☕