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 will use a car exactly one time in a day if she selects at random is 4/9 or 0.44

Step-by-step explanation:

Extracting the key information from the question:-

** * Kathy is faced with 3 transportation options daily.

** * She can either take a bus, car or train to work in the morning.

** * In the evening she is also faced with the same options.

** * We are required to find/calculate the probability that she uses a car exactly one time if she randomly chooses her mode of transportation.

Now, we are told that Kathy can always make use of any of the three vehicles to get herself transported to work and back home. The probability of selecting any of those vehicles will be 1/3. This means that the probability of using a car = 1/3. The probability of using a train is 1/3 and the probability is also the same for a bus.

Next is to calculate the probability that she uses a car exactly one time. This means she can only make use of a car just once a day (either when going to work or while returning from her workplace). If the probability of using a car = 1/3 then the probability of using any other vehicle apart from a car will be = 2/3 (that is the combined probability of us and train = 1/3 + 1/3).

The probability of using a car only once = probability of using it in the morning + the probability of using it in the evening.

That is P (using a car exactly once) =

[ (P (car) * P (others)) + (P (others) * P (car) ]

P (using a car exactly once) =

(1/3 * 2/3) + (2/3 * 1/3)

= 2/9 + 2/9

= 4/9

Therefore, the probability that Kathy will make use of a car exactly once in a day is = 4/9 or 0.44

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.