Elizabeth lives in San Francisco and works in Mountain View. In the morning she has 3 transportation options (take a bus, a cab, or a train) to work, and in the evening she has the same 3 choices for her trip home. If Elizabeth randomly chooses her ride in the morning and in the evening, what is the probability that she’ll use a cab exactly one time.

There are nine possible combinations as follows: bb, bc, bt, cb, cc, ct, tb, tc, tt, each having a probability of 1/9. The four combinations with a cab exactly one time are: bc, cb, ct, tc. Therefore the probability that she'll use a cab exactly one time is: 4/9.

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psm22415 psm22415 · Answer 2 · 2022-05-08T14:29:20+02:00

The probability that Elizabeth uses a cab only once is 4/9.

What is the probability?

Probability is a branch of math that deals with finding out the likelihood of the occurrence of an event.

Elizabeth lives in San Francisco and works in Mountain View.

In the morning she has 3 transportation options (take a bus, a cab, or a train) to work, and in the evening she has the same 3 choices for her trip home.

All transportation (bus, cab, train) are all similarly likely to be selected, and 1 of them must be selected in the morning and evening, so we get:

P (bus) = P (cab) = P (train) = 1/3.

We also have P(no cab in evening) = P(no cab at morning) = 2/3

Now, P(using cab exactly once) = P(cab at morning and no cab in the evening) + P(no cab at morning and cab in the evening)

= P(cab, no cab) + P(no cab, cab)

= 1/3 × 2/3 + 2/3 × 1/3

= 2/9 + 2/9

= 4/9

Hence, the probability that Elizabeth uses a cab only once is 4/9.

Learn more about probability here;

https://brainly.com/question/10590289

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