Sample space with equally likely outcomes: question to a worked out example

Question

Attached is an extract from the book A First Course in Probability by Sheldon Ross. I am struggling to understand the flow of his logic at the place highlighted in red.

For example, let $n = 6$, $k = 4$. Then there are 4 possible events $A_i$ with $i=1,2,3,4$ provided we do not distinguish among non-special balls, i.e. their order does not matter. Also, in case of each event we end up with 4 balls. How does the author come to the conclusion that $P(A_1) = P(A_2) = P(A_3) = P(A_4) = 1/6$? I do not understand what he means by “since each one of the n balls is equally likely to be the $i$th ball chosen”.

Is there a way to come to this conclusion systematically? If so, how does it look like? Or could you help me please to understand how the author meant it? Thanks.

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Answer 1

The probability of drawing any particular ball is the same as drawing another, as stated at the beginning of the exercise.

Call this probability $p$ and say there are $6$ balls in the urn. Since the probabilities are the same and we know all probabilities in the sample space must add to $1$ we have

$$p+p+p+p+p+p=1$$

$$6p=1$$

$$p=\frac16$$

This works for any number $n$ of balls in the urn:

$$\sum_{i=1}^np=1$$

$$np=1$$

$$p=\frac1n$$

Answer 2

$A_i$ refers to the event of the special ball being chosen on the $i^{\text{th}}$ draw. Meaning, $A_1$ is the event in which you choose the special ball on your first draw. Similarly, $A_4$ is the event in which you choose the special ball on your fourth draw.

Now, what is the probability, the probability that the first ball you choose is the special ball?

In your example, there are $6$ balls, each equally likely to be chosen. So,

$$P(A_1) = \frac{1}{6}$$

Easy, right? Now, what is the probability of $A_{4}$ - the special ball is chosen on the $4^{\text{th}}$ draw?

This might not appear as straightforward. You might go with something like this.

For the fourth ball to be the special ball, the first, second, and third ball must be ordinary ones and then the fourth must be special. So,

$$\begin{align*}P(A_4) &= \frac{5}{6} \times \frac{4}{5} \times \frac{3}{4} \times \frac{1}{3} \\[0.3cm] &= \frac{1}{6}\end{align*}$$

See? $1/6$ again.

And we didn't even have to do all this work. There is an easier approach - one that the author is proposing.

Go back to your example, forget all the calculations we have done so far, and again what is the probability of choosing the special ball on the fourth draw ($A_4$)?

There is a symmetry in the situation. Each ball (including the special one) is the same as far as our choices are concerned. So, the probability of choosing the special ball on the fourth draw is the same as choosing any other ball.

And there are a total of $6$ balls. So, the probability must be $1/6$.

You can generalize it. The probability of choosing any particular ball on any particular attempt (sixth or less) would be the same - $1/6$. Simply because every ball is equally like to be selected and hence indistinguishable in that aspect.$^{[1]}$

Generalizing it to $n$ balls, the probability $P(A_1) = 1/n$. That's what you see in the book.

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$^{[1]}$ More explanation in response to OP's comments below

Symmetry plays a very important role in the calculation of probabilities. Two symmetrical events must have the same probability (this may not be the most technically accurate language though). This helps us find probabilities easily with limited/no additional use of permutations and combinations.

Consider the third (or any of the four) draw. What are the different possibilities? The chosen ball must be either the special one ($E_S$) or one of the five ordinary ones ($E_{O_1}$, $E_{O_2}$, $E_{O_3}$, $E_{O_4}$, or $E_{O_5}$). No other event is possible.

Now ask yourself, is there anything associated with any of the balls that makes it more or less likely to be drawn? The question clearly says they are equally likely.

But also by intuition. The balls are identical (except for the color, and you are drawing without looking at the color). So why (or how) should any ball have a greater or smaller probability to be selected?

Moving forward, if they are equally likely, it means the individual probabilities for each of the six events are equal (say $p$). Also, together they must add to 1 (together the events cover every possibility).

$$\begin{align*} 6p&= 1\\[0.3cm] \Longrightarrow \,\,\, p &= \frac{1}{6}\end{align*}$$