# Rental truck probability confusion

would appreciate if anybody could help me with this probability problem

A rental truck agency services its vehicles on a regular basis, routinely checking for mechanical problems. Suppose that the agency has six moving vans, two of which need to have new brakes. During a routine check, the vans are tested one at a time.

a. Given that one van with bad brakes is detected in the first two tests, what is the probability that the remaining van is found on the third or fourth test?

This solution was provided by the text book but it doesn't look right to me, I am having hard time understanding it if some one can elaborate it, I'll really appreciate it.

In order that no more than four vans must be tested, you must find one or both of the faulty vans in the first four tests. choosing four from the six vans, and P(one or two faulty vans) = Isn't this a conditional probability, so shouldn't it be divided by some other number rather than total possible outcomes.

There are $$\binom{6}{2}=15$$ possible and equally likely ways in which two faulty vans can be situated among the sequence of six vans. The graphic depicts them by plotting each possible configuration as as vertical column of cells, shaded to depict where the two faulty vans might be located.
Only in the eight configurations $$2$$ through $$9$$ is exactly one faulty van found in the first two tests.
In half of these configurations (namely, configurations $$2,$$ $$3,$$ $$6,$$ and $$7$$) the second faulty van appears in the third or fourth position. I hope this is obvious, because conditional on there being just one remaining van in positions three through six, it is equally likely for that van to be in the next two or last two positions.
Therefore the solution is $$4/8 = 1/2,$$ not $$9/15$$ as claimed.