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Are the elementary events:

4 because you'll either pick a red, green, white or yellow ball?

or

are they 7 because there are 3 red, 2 green, 1 white and 1 yellow balls?

Also, if the question is " Box contains 3 red, 2 green, 1 white and 1 yellow balls. You draw 1 ball, remember its color and return it to the box, then you draw another one. How many elementary events are there?" does that mean there will be even more elementary events because we'd to take into consideration the second drawn ball?

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  • $\begingroup$ As long as the balls are indistinguishable (except that for the colour) the events are 4 with different probabilities (3/7, 2/7, 1/7, 1/7). For the second problem the events are 16. $\endgroup$ – N74 Jun 20 '18 at 19:07
  • $\begingroup$ You could argue that there are even more elementary events... For example "the ball drawn was red and the neighbors are cooking pork tonight", "the ball drawn was red and the neighbors are cooking chicken tonight", etc... (assuming the neighbors cook at most one meat each night) You can make the sample space as needlessly complicated or as simple as you like. The only requirement is that every event you wish to discuss can be represented by the union of these elementary events, the elementary events don't overlap, and the union of all elementary events is the sample space. $\endgroup$ – JMoravitz Jun 21 '18 at 1:20
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It basically means counting all possibilities... such as if you take out say a red ball what other events can happen in tandem with this, creation of the sample space of all outcomes is probably the goal here... hope this helps.

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An elementary event is an event containing only a single outcome; an atom.

The outcomes of the first trial are results for a single drawing.

The outcomes for the second trial are ordered pairs of the results of each from the two draws (with replacement) — therefore you must count the distinct pairs.

How many there are to count depends on whether you consider the balls of the same colour to be distinctive.

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