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I'm trying to create a sample space of elementary events for the following problem, however am having difficulty. Does anyone know how to create this set of elementary events?

In a factory there are 100 units of a certain product, 5 of which are defective. We pick three units from the 100 units at random. What is the probability that none of them are defective?

Let us define $A_i$ as the event that the $i$th chosen unit is not defective, for $i = 1, 2, 3$. We are interested in $P(A_1 \cap A_2 \cap A_3)$.

I understand this is a conditional probability problem with the solution $95/100 * 94/99 * 93/98$. My initial attempt at the sample space is $E = \{U_1,U_2 ... U_{100}\} x \{U_1,U_2 ... U_{99}\} x \{U_1,U_2 ... U_{98}\}$ which yields the correct answer. I'm looking to create a more comprehensive sample space based on the permutations of the set of units when removing items. For instance, the above solution is simply comparing a sample size count to the probability space, without taking into account permutations of the units.

My approach was to use the cartesian product over multiple sets considering all the permutations. For instance, the possibilities of removing 1 item from 100 is $100 \choose 99$. The confusing part is $99 \choose 98$ will be a different set for each of the former. I keep getting stuck trying to create this and some elementary events. Guidance appreciated.

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2 Answers 2

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The elementary events of your initial attempt at a sample space are triples, $\ \big(U_{i_1}, U_{i_2}, U_{i_3}\big)\ $ of units in which $\ 1\le100, 1\le i_2\le99\ $ and $\ 1\le98\ $. This isn't quite right, because it includes triples such as $\ \big(U_5, U_5, U_6 \big)\ $, in which the same unit, $\ U_5\ $, has been chosen twice, and it doesn't include triples such as $\ \big(U_1, U_{100}, U_{99}\big) $, where the second unit chosen is $\ U_{100}\ $ or the third unit chosen is $\ U_{99}\ $ or $\ U_{100}\ $. Your description of the problem, however, implies that the sampling is done without replacement—that is, the three units chosen must be distinct—and the second and third units chosen could be any one of the $100$ units.

You also need to distinguish the defective units from the non-defective ones, which you could do by choosing $5$ of your indices—$1$ to $5$, for example, or $96$ to $100$—to be those of the defective units.

In setting up the event whose probability you're looking for, you've chosen to distinguish the units by the order in which they're chosen. This isn't really necessary for this particular problem, but it's nevertheless a perfectly sound approach. If you let $\ A(1)\ $, $A(2)\ $, and $A(3)\ $ be the first, second and third units chosen, then $\ A:\{1,2,3\}\rightarrow$$\big\{U_1,U_2,\dots,U_{100}\big\}\ $ is a function which must have the property that $\ A(1), A(2), A(3)\ $ are all distinct—that is, it must be one-to-one. So you could take your sample space to be $$ \Omega=\big\{A:\{1,2,3\}\rightarrow\big\{U_1,U_2,\dots,U_{100}\big\}\,\big|\,A\ \text{is one-to-one}\big\}\ . $$ This sample space contains $\ 100\cdot99\cdot98\ $ elementary outcomes, all of which are equally likely, and if you take $\ U_{96}, U_{97}, \dots, U_{100}\ $ to be the defective units, then the event whose probability you're looking for is $$ \big\{A\in\Omega\,\big|\,A(1),A(2),A(3)\in\big\{U_1,U_2,\dots,U_{95}\big\}\big\}\ . $$ This event comprises $\ 95\cdot94\cdot93\ $ elementary outcomes, each of probability $\ \frac{1}{100\cdot99\cdot98}\ $, so its probability is $\ \frac{95\cdot94\cdot93}{100\cdot99\cdot98}\ $, as you've already found.

The reason why it's not really necessary to distinguish the order in which the units are chosen is that the occurrence or non-occurrence of the event whose probability you're interested in doesn't depend in any way on the individual values of $\ A(1), A(2), A(3)\ $ separately, but merely on whether the set of them, $\ \big\{A(1),A(2),A(3)\big\}\ $ is a subset of $\ \big\{U_1,U_2,\dots,U_{95}\big\}\ $ or not, so you can simplify your sample space a little by taking it to be the collection of $3$-element subsets of $\ \big\{U_1,U_2,\dots,U_{100}\big\}\ $: $$ \Omega=\big\{A\subseteq \big\{U_1,U_2,\dots,U_{100}\big\}\,\big|\,|A|=3\,\big\}\ . $$ This sample space comprises $\ {100\choose3}\ $ elementary outcomes, again all equally likely, and the event whose probability you're interested in is now $$ \big\{A\subseteq \big\{U_1,U_2,\dots,U_{95}\big\}\,\big|\,|A|=3\,\big\}\ , $$ Which contains $\ {95\choose3}\ $ elementary outcomes, each of probability $\ \frac{1}{100\choose3}\ $, so its probability is $\ \frac{95\choose3}{100\choose3}\ $, with the same value as before.

Response to OP's comment

Your sample space can consist of any mathematical objects you like, as long as the model you construct is a faithful representation of the possible outcomes that can occur in reality. On reflection, given the approach you had already taken, I think my use of functions to represent the outcomes wasn't the best choice, since a small modification of your own approach would have sufficed. You can simply take your sample space $\ \Omega\ $ to be given by $$ \Omega=\big\{\big(A_1,A_2,A_3\big)\in{\cal U}^3\,\big|\,A_1\ne A_2\ne A_3, A_1\ne A_3\big\}\ . $$ where $\ {\cal U}= \big\{U_1,U_2,\dots,U_{100}\big\}\ $.

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  • $\begingroup$ Thanks, the reason for my first approach was to show conditional probability of chained events which was the problem tutorial. Unsure what that would look like using your approach. It looks like I can use functions as elementary events, which I didn't know before. I was hung up on creating a sample space that included all the unordered arrangements of the remaining good units after each selection of good unit. Also, the same for good - defective, however that sounds like it's too complicated. $\endgroup$
    – Nick
    Commented Oct 20, 2020 at 2:57
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If the units are $U_1, ..., U_{100}$ then the best/typical way of representing the sample space is $\{ (U_a, U_b, U_c): 1 \leq a < b < c \leq 100 \}$, the set of all different groups of $3$ units without counting order. This is a nice way of thinking about it because it makes it easy to generalize the probability of getting exactly $k$ defectives, $k = 0, 1, 2, 3$, in 3 draws: $$P(k \text{ defectives }) = \frac{\binom{95}{3-k} \binom{5}{k}}{\binom{100}{3}}.$$ We can check that this agrees with the answer $95/100 * 94/99 * 93/98$ when $k = 0$.

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