Honestly, sometimes my first off-the-cuff measure of how well we do collectively with a topic in mathematics is how well I do with it. This is terrible reasoning, of course, but it has some uses as a kind of first measurement that needs to be independently verified. It helps me notice potential weak spots in instruction, anyway, like with independent events.

The concept of independent events is certainly a candidate for being a weak spot. Most of what I’ve seen online doesn’t really crack the surface. Students are allowed to explore the concept of independent events, but all that seems to mean is to take a situation and tell me whether the events are independent. Big whoop.

And the definition often used—that two events are independent if the occurrence of one does not affect the probability of the other—has the potential of really confusing students (and me). Consider this situation.

Are the events “selecting a letter card” and “selecting a circle card” independent events? The way a student (and I) might reason, given just the definition above, would be to think, “Well, if I pick B, that would definitely change the probability of picking a circle, because the B-card is also a circle card. And, if I pick A, that would change the probability of picking a circle card, because then I would only have three cards to choose from. So, the events are not independent.”

Try This Instead

The above was perfectly sane reasoning; it’s just wrong because of a terrible explanation (or lack of an explanation in most cases). I thought of something maybe a little better. Here it is in sentence form, similar to how we worked on the impenetrability of trig ratios:

You have the same probability of choosing a letter card from all the cards as you have of choosing a letter card from just the circle cards.

This is what makes choosing a letter card and choosing a circle card “independent.” If I **know** that I’ve drawn a circle, the probability that I’ve also drawn a letter, \(\mathtt{0.5}\), is the same as if I didn’t know I’d drawn a circle. And of course this works automatically the other way around too: if I know that I’ve drawn a letter card, the probability of drawing a circle is the same as if I didn’t know. **If I reduce the sample space from all cards to circle cards, the probability of “letter” is the same.**

At the heart of independent events (besides the conditional probability flavoring above) are equivalent ratios, or a proportion . . . which gives me an ultra-short way of saying it a little more mathematically:

\[\mathtt{\frac{2}{4} = \frac{1}{2}}\]

Or, “2 letter cards out of 4 cards in all is the same as 1 letter card out of 2 circle cards.” In the symbolism of probability, we actually write this with complex fractions (by giving each numerator and denominator above a denominator of 4), and then disguise the complex fractions with \(\mathtt{P()}\) statements, which is all equivalent to the above proportion:

\[\mathtt{\frac{\color{purple}{\frac{2}{4}}}{\color{red}{\frac{4}{4}}} = \frac{\color{red}{\frac{1}{4}}}{\color{purple}{\frac{2}{4}}} \longrightarrow \frac{\color{purple}{P(\textrm{letter})}}{\color{red}{1}} = \frac{\color{red}{P(\textrm{letter and circle})}}{\color{purple}{P(\textrm{circle})}}}\]

And that gives us the definition of independence using conditional probability, from S-CP.A.3. If we remember our proportion work from way back when, then the “other” test for the independence of two events pops out of the equivalence of the products of means and extremes: \[\mathtt{P(\textrm{letter}) \cdot P(\textrm{circle}) = P(\textrm{letter and circle})}\]

The complex fraction part of this explanation seems to be the most important, actually. And we don’t really do a good job of letting kids in on that disguise either. But still, stapling the idea of independent events to a pair of equivalent ratios (a proportion) helps the whole idea make a lot more sense to me. And, truthfully, it makes the notion of “independence” as “not having an effect on another probability” seem almost wrong.

**Update:** This kind of reasoning works for the typical example of independent events. The situation involving separate spinners is fairly easy for kids to identify as being about independent events, but like a lot of other topics in mathematics education, we start off with examples that are easy and also completely misleading. Then we all opine that kids have difficulties because the material gets “harder.” Anyway, spinning a C on the first spinner and a 2 on the second spinner are independent events, but not because there are “independent” spinners.

What’s the proportion (if the events are independent) that matches the situation?

Assuming we spin the first spinner and don’t know what we get, there are \(\mathtt{3 \times 1}\), or 3, outcomes that have 2 as the second spin, out of 12 possible outcomes. The outcomes are {(A, 2), (B, 2), (C, 2)}. But, if we know that we have spun a C first, then there is 1 outcome showing 2 on the second spinner, out of 4 possible outcomes. So, our proportion is \[\mathtt{\frac{3}{12} = \frac{1}{4}}\]

This is all we need to show that the two events are independent, actually. If that proportion is true, then the events are independent. But we can cue the complex fraction magic again for reinforcement: \[\mathtt{\frac{\color{purple}{\frac{3}{12}}}{\color{red}{\frac{12}{12}}} = \frac{\color{red}{\frac{1}{12}}}{\color{purple}{\frac{4}{12}}} \longrightarrow \frac{\color{purple}{P(2)}}{\color{red}{1}} = \frac{\color{red}{P(\textrm{C and 2})}}{\color{purple}{P(\textrm{C})}}}\]