Assume that there are 7 cars: 3 Chevrolets, 3 Fords, and 1 Toyota. Two cars leave at random, one after the other. Draw a tree diagram and use it to answer the following questions.
(1) What is the probability that the Toyota leaves first?
(2) What is the probability that the two cars which leave are Chevrolets?
(3) What is the probability that the Toyota is the second car to leave?
(4) What is the probability that a Ford is the last to leave given that the Toyota
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) If the Toyota is the first to leave, then it can't leave second.
So the probability that the Toyota is the second car to leave is
P(not Toyota from 9) x P(Toyota from 8)
8/9 x 1/8 = 8/72 = 1/9
4) There are three possibilities:
P(Toyota from 9) x P(Any other car from 8) = 1/9 x 8/8 = 8/72
P(Not Toyota from 9) x P(Toyota from 8) = 8/9 x 1/8 = 8/72
P(Not Toyota from 9) x P(Not Toyota from 8) = 8/9 x 7/8 = 56/72
These all add up to 1, so we've got them correct.
Now, we're only interested in the ones where the Toyota is included, i.e. the first two.
And their probabilities are equal, so the answer is 1/2
5) I assume "last to leave" means "second to leave", not "ninth to leave"!
Given that the Toyota leaves first, then you already know the makeup of the cars for who is second to leave: 5/8
thats not right. 3 people have copied and pasted that from someone else who tried to help me