200 cars, 6 defective; probability of only the last in a selection of four defective 
A consumer agency claims that in 200 units of a newly launched model of a car, six units of such car faced a brake system problem. If 4 cars are randomly chosen, find the probability that only the last car selected will face the problem.

The given answer is 0.1111 but of course answers given are not always correct. I am just a normal student learning maths in my secondary school and one student proposed this answer:
$$\frac{\binom{194}3\binom61}{\binom{200}4}$$
Another student proposed this answer:
$$\frac{194}{200}\cdot\frac{193}{199}\cdot\frac{192}{198}\cdot\frac6{197}$$
This one is different from the answer scheme. The teacher said that the second answer is true as this question wants only the last brake to be defective. Why is the first one wrong?
 A: The second answer is correct.  Without loss of generality we can number the vehicles from $1$ to $200$ such that the first six vehicles have the defect, and the rest do not.  We can reason that if the sample is $(X_1, X_2, X_3, X_4)$, where $X_i$ represents the number of the $i^{\rm th}$ sampled car, then $$\Pr[(X_1, X_2, X_3 > 6) \cap (X_4 \le 6)] \\ = \Pr[X_1 > 6]\Pr[X_2 > 6 \mid X_1 > 6]\Pr[X_3 > 6 \mid X_1, X_2 > 6]\Pr[X_4 \le 6 \mid X_1, X_2, X_3 > 6] \\
= \frac{194}{200} \cdot \frac{193}{199} \cdot \frac{192}{198} \cdot \frac{6}{197},$$ as claimed.
The reason why the first method fails is because the binomial coefficients do not take into account the order in which the cars are sampled, therefore, there is no way the resulting calculation properly counts the desired event, since the desired event involves observing three non-defective cars followed by a defective car.  What does it count, then?  The expression counts the probability that, if four cars are selected at random without replacement, there will be exactly one car with the defect--but not caring whether the defective car was the last one in the sample.  You will note that since there are $4$ permutations of three good cars and one defective car, the value $$\frac{1}{4} \frac{\binom{194}{3}\binom{6}{1}}{\binom{200}{4}}$$ is equivalent to the answer given by the  second method.
A: The first answer is wrong because it is the probability of selecting exactly one problematic car out of four without regards to order. (It is in fact a value of the hypergeometric distribution $\operatorname{hgeom}(200,6,4,1)$). The question says "last", so there is an order, which leads to the very simple but correct calculation of the second answer.
A: Given that 6 out of 200 cars are defective, the probability the first car chosen is NOT defective is 194/200= 97/100.  Given that this does happen, there are 6 defective cars out of 199.  The probability the second car chosen is not defective is 193/199.  Given that this does happen, there are 6 defective cars out of 198.  The probability the third car chose is not defective is 192/198= 96/99.  Given that this does happen there are 6 defective cars out of 197.  The probability the four car chose IS defective is 6/197.
The probability only the fourth car is defective is (97/100)(193/199)(96/99)(6/197), about 0.563%.
