[Math] {students 1 and 2 are in different groups} vs {students 1, 2, 3, and 4 are in different groups}

probabilityproof-writing

Source: Example 1.11, p 26, *Introduction to Probability (1 Ed, 2002) by Bertsekas, Tsitsiklis.

Hereafter abbreviate graduate students to GS and undergraduate students to UG.

Example 1.11. A class consisting of 4 graduate and 12 undergraduate students
is randomly divided into 4 groups of 4. What is the probability that each group
includes a GS? We interpret “randomly” to mean that given the
assignment of some students to certain slots, any of the remaining students is equally
likely to be assigned to any of the remaining slots.

Solution: We then calculate the desired
probability using the multiplication rule, based on the sequential description shown
in Fig. 1.12. Let us denote the four GS by 1, 2, 3, 4, and consider
the 4 events
$A_1$ = {GS 1 and 2 are in different groups},
$A_2$ = {GS 1, 2, and 3 are in different groups},
$A_3$ = {GS 1, 2, 3, and 4 are in different groups}.

We will calculate $\Pr(A_3)$ using the multiplication rule:

$\Pr(A_3) = \Pr(A_1 ∩ A_2 ∩ A_3) = \Pr(A_1)\Pr(A_2 |A_1) \Pr(A_3 \mid A_1 \cap A_2)$. $\qquad […]$

  1. How is $\Pr(A_3) = \Pr(A_1 ∩ A_2 ∩ A_3)$?

  2. I understand that the question asks for $\Pr(A_1 ∩ A_2 ∩ A_3)$; but how would you know (or divine) to reinterpret and then rewrite $\Pr(A_1 ∩ A_2 ∩ A_3)$ as $\Pr(A_3)$? 1 appears the key but tricky but step in formulating this problem.

Best Answer

  1. How is $\Pr(A_3) = \Pr(A_1 ∩ A_2 ∩ A_3)$?

If GS 1, 2, 3, and 4 are in different groups, then

  • GS 1, 2, and 3 are in different groups; and
  • GS 1 and 2 are in different groups.

Therefore, if $A_3$, then $A_2$ and $A_1$.

In set-theoretic terms, $A_3 \subseteq A_2 \subseteq A_1$. The intersection of a set with any one of its subsets is the subset. Therefore, $A_1 \cap A_2 \cap A_3 = (A_1 \cap A_2) \cap A_3 = A_2 \cap A_3 = A_3$.

  1. I understand that the question asks for $\Pr(A_1 ∩ A_2 ∩ A_3)$; but how would you know (or divine) to reinterpret and then rewrite $\Pr(A_1 ∩ A_2 ∩ A_3)$ as $\Pr(A_3)$? 1 appears the key but tricky but step in formulating this problem.

The motivation behind recasting $\Pr(A_3)$ as $\Pr(A_1 \cap A_2 \cap A_3)$ is a desire to work with conditional probabilities, such as the ones in $\Pr(A_1) \Pr(A_2 \mid A_1) \Pr(A_3 \mid A_1 \cap A_2)$. In this light, $\Pr(A_1 \cap A_2 \cap A_3)$ acts a bridge to conditional probabilities.

When you ask how I would know to recast the problem in terms of conditional probabilities, I assume you mean how should you have known. In short, you shouldn't have known! You encountered the problem as an example in a textbook. The purpose of the example is to teach you how to solve problems of this sort. You were expected not to know the method of solution beforehand. (If, to the contrary, you often do know the method of solution beforehand, then you are reading below your level.)

Take this example as a lesson: The relationship between probabilities of intersections and conditional probabilities can be a useful one to exploit when encountering either. Now, when you encounter a similar problem in the exercises, you should know to consider recasting the problem in terms of conditional probabilities.

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