All transportation (bus, cab, train) are all similarly likely to be selected, and 1 of them must be selected at 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
Probability that Elizabeth uses a cab only once is 4/9.
What is the area of the triangular base?
8 square feet
What is the length of edge DF?
4 feet
Answer:

Step-by-step explanation:
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