3 vehicles out of 5 randomly showcased, what is the probability of each vehicle being showcased? I have a basic probability doubt. If I have 5 different vehicles backstage, out of which 3  random can be showcased to the public. What is the probability of each vehicle to get to showcase? Once a car is in a showcase, it will not be returned to backstage.
I thought by doing the calculating the combinations- ${5 \choose 3}=10$.
And probability of 3 slots getting filled (let's say $P$) - $P=\frac{1}{5}\times\frac{1}{4}\times\frac{1}{3}$.
So, total probability is $10\,P= \frac{1}{6}$.
Or will it be $\frac{3}{5}$?
Or any other solution?
 A: You asked for alternative approaches.  Here is one you might find useful.
Let's begin by stating the obvious: you are implicitly assuming the five probabilities are equal.  The expected total in the showcase equals the sum of those probabilities, whence it is five times any one of them.  Yet the expected total is the average value of all possible totals, weighted by their chances of occurring.  Since by design the possible total is always $3$, its average must be $3$.  Therefore each probability is $3/5$.
The power of this reasoning about expectations becomes clear when you generalize the question to $k$ cars in the showcase to be chosen (with equal probabilities) out of $n$ cars backstage.
A: I assume that all cars are equally likely to be chosen.
Suppose without loss of generality the cars are labeled 1 through 5. The probability of not choosing car 1 is (4/5) * (3/4) * (2/3) = 2/5, so the probability of choosing it is 1 - 2/5 = 3/5. Of course, by my assumption above, the same argument applies to any of the cars. The answer 3/5 makes intuitive sense since we're drawing 3 things from a set of 5.
