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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?

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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.

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  • $\begingroup$ But what about the possible combinations of C1,C3,C4 or C2,C5,C1 etc. Will the 3/5 probability have that into account as well? $\endgroup$ – Joe Jul 1 '16 at 18:55
  • $\begingroup$ You can renumber the cars however you like and the argument is the same. $\endgroup$ – Kodiologist Jul 1 '16 at 18:56
  • $\begingroup$ Thank You! Can you please tell in which scenario will my logic below (the same i described in the question) be used <br>- 5C3=10 And probability of 3 slots getting filled (let say P)- 1/5*1/4*1/3 So, total probability = 10*P= 1/6 $\endgroup$ – Joe Jul 5 '16 at 13:36
  • $\begingroup$ If we have 60 Cars backstage in place of 5, will he situation become- Probability of not chosing car1 is (59/60)*(58/59)*(57/58) = 57/60, so the probability of chosing it will become 1-(57/60)= 3/60 $\endgroup$ – Joe Jul 7 '16 at 13:51
  • $\begingroup$ Yes, that's right. $\endgroup$ – Kodiologist Jul 7 '16 at 14:45
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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.

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