Kathy lives in Dallas and works in Woodrow Wilson. In the morning, she has 3 transportation options (take a bus, a car, or a train) to work, and in the evening, she has the same 3 choices for her trip home. If Kathy randomly chooses her ride in the morning and in the evening, what is the probability that she will use a car exactly one time?

The probability that Kathy uses the car only once a day is 4/9.

Step-by-step explanation:

Let's recall that the formula of the probability is:

P = Number of favorable outcomes/Total number of possible outcomes

In consequence, Kathy can randomly select any transportation option (bus, cab, train) in the morning and in the evening,

So we have:

P (bus) = P (car) = P (train) = 1/3. 

We also have:

P (no car in the morning) = P (no car in the evening) = 2/3 

For finding the probability that Kathy will use the car exactly once a day, we calculate it this way:

P (using car exactly once) = P (car in the morning and no car in the evening) + P (no car in the morning and car in the evening)

Replacing with the values we know:

P = (1/3 * 2/3) + (2/3 * 1/3) 

P = 2/9 + 2/9 

P = 4/9 

The probability that Kathy uses the car only once a day is 4/9.

y= -