Venn Diagrams and Formulas

At the math office today, two math teachers were discussing probability. Two things were surprising about it. One, it was a Saturday afternoon part of our winter break, so no teachers should have been on campus. (We’re a dedicated lot, us math teachers.) Two, the topic they were discussing was so simple, and yet, it reveals the real mind-bending character that probability has on us and our students.

Question 1: What is the probability that you draw a heart or a queen from a deck of cards?
Question 2: What is the probability that you roll a die and get a number less than 4 or an odd number?

Both questions are simple enough. The first one is 16/52 (because there are 16 cards which are hearts or queens in a deck). The second one is 4/6 (because you can roll a 1, 2, 3, or 5).

The problems are seemingly the same. Let’s now look at this problem from the perspective of a venn diagram.

picture-1If we want to know the probability that event A or event B occurs, we clearly can see that we have:

P(A\text{ or }B)=P(A)+P(B)-P(A \text{ and } B)

(We have to subtract that last term, because we added that overlapping section twice when we took P(A)+P(B).)

Let’s apply that to our two questions:

Question 1: We have a probability of: P(\text{Heart or Queen})=P(\text{Heart})+P(\text{Queen})-P(\text{Heart and Queen}). Clearly P(\text{Heart})=13/52 and P(\text{Queen})=4/52. What is P(\text{Heart and Queen})? Intuitively — or using our venn diagram — we know it is 1/52. And the answer works out correctly to 16/52.

Question 2: We have a probability of:  P(\text{less than 4 or odd number})=P(\text{less than 4})+P(\text{odd number})-P(\text{less than 4 and odd number}). Clearly P(\text{less than 4})=3/6 and P(\text{odd number})=3/6. What is P(\text{less than 4 and odd number})? Intuitively — or using our venn diagrams — we know it is 2/6. And the answer works out correctly to 4/6.

However, let’s say we wanted to calculate P(\text{Heart and Queen}) and P(\text{less than 4 and odd number}) (the overlapping regions) mathematically? It turns out that there is something fundamental that makes these two problems different. In question 1, the two events (drawing a heart / drawing a queen) are independent. In question 2, the two events (rolling a number less than 4 / rolling an odd number) are dependent. For the first question, you can say that P(\text{Heart and Queen})=P(\text{Heart})P(\text{Queen}) while in the second problem you cannot do that.

Recall that the definition of independence of two events A and B is if P(A|B)=P(A).

Checking the first question for independence, we see that the probability of drawing a heart given that you already have a queen is 1/4, and that is the same as the probability of drawing a heart (1/4). (Similarly, the probability of drawing a queen given that you already have a heart is 1/13, and that is the same as the probability of drawing a queen (1/13).) So the two events are independent.

Checking the second question for independence, we see that the probability of rolling and odd number given that you have rolled a number less than 4 is 2/3, while the probability of rolling an odd number is 1/2. (Similarly, the probability of rolling a number less than 4 given that you’ve rolled an odd number is 2/3, while the probability of rolling a number less than 4 is 1/2.) So the two events are dependent.

The teacher who brought up this problem was grading exams, and one student had calculated P(\text{Heart and Queen})=P(\text{Heart})P(\text{Queen}). And seeing the two problems were almost identical, calculated P(\text{less than 4 and odd number})=P(\text{less than 4})P(\text{odd number}) — which, as we know, isn’t right for dependent events.

What we were discussing is how we could explain to the student that the two situations are different, even though on the surface the questions seem like they are of the same form. In other words, is there a conceptual — non mathematical — way to explain that the first question involves independent events while the second question involves dependent events? It certainly isn’t intuitive, at least not to me.

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14 comments

  1. Maybe it has to do with the idea that cards have two properties that are unrelated “value” and “suit.” Those are clearly independent (although something like “color” is dependent on suit).

    Die just have a single property: “value” (or “number”).

    So, the first question is asking about two different properties while the second is asking about the same property in two different ways.

  2. One way I try to get the idea of independent events across is with a simple card illustration. I draw one card, hold it facing me, and ask, “What is the probability that this is a Heart?” “One fourth.” “Good. Now, if I were to tell you that this card is an Ace, would that change the probability that the card is a Heart to something other than one fourth?” (pause…) “No.” “Right. Now, what’s the probability that this card is an Ace?” “One thirteenth.” “Right. And if I were to tell you that this card is a Heart, would that change the probability that the card is an Ace to something other than one thirteenth?” “No.” “Right. The fact that our knowledge of one event has no effect on the probability of the other means that these events are independent of each other.”

    I’m not sure that’s as non-math as you’d like, but one reason I like it is that it kind of meshes with the students’ preconceived notion of the word “independent”.

  3. I wouldn’t have them multiplying probabilities at all for these two problems. I keep probability as simple as possible because of the groups I teach it to (9th graders who are not honors or accelerated math students).

    These are single events…so I’d have the kids doing (number of Queens of Hearts)/(total) and (number of odds less than 4)/(total).

    I talk about dependent and independent when calculating probabilities of consecutive events, wherein “without replacement” implies dependent, so you have to adjust values in the second probability accordingly before multiplying. (As in, “what is the probability of drawing a queen and then a heart, without replacement”.)

  4. @Dave, I’m not sure that the two unrelated properties is right… because they aren’t unrelated. Maybe because EVERY combination of the two properties is included that it works.

    @Matt, I think that’s spot on. I’ll suggest that to the teacher. I guess there is nothing on the surface to suggest they are independent/dependent events. Like when you draw cards/balls without replacement, you KNOW that they are dependent events. But I guess in your head you’re intuitively doing the calculation, or at least KNOW that one will affect the other. You don’t know that for the second question without doing the calculation.

    @Kate, totally agreed! I think the teacher assumed that all students would solve it with sample spaces. This kid (precalc) solved it using the formulas, though. Which got us math teachers thinking about how two seemingly similar problems actually had a fundamental difference between them.

    @Jackie, yes! I’ll suggest that to the teacher.

    Thanks y’all.

  5. I know we like to move our students beyond rote formula memorization, but I like the fact that a lot of probability can essentially be boiled down to a couple of formulas, one for “or”, which you mentioned in your post, and one for “and”, which the student did not get quite right.

    P(A or B) = P(A) + P(B) – P(A and B)

    (If A and B are mutually exclusive, then P(A and B) = 0, making the formula P(A or B) = P(A) + P(B).)

    P(A and B) = P(A) P(B|A) = P(B) P(A|B)

    (If A and B are independent events, then P(B|A) = P(B) and P(A|B) = P(A), and the formula becomes P(A and B) = P(A) P(B).)

    1. hola, me podrias todas las formulas por favor, me urge, cuando son excluyentes, binomial etc.
      como sacar a unido ab, a interceptado a B, a dado que b.. por favor

  6. Heh, I actually had one for three years, from ’01 to ’04, though it wasn’t specifically about teaching. I’ve thought about getting back into it, but I feel pressed for time as it is!

  7. @matt: Bah humbug! LAME! You should get it, and write in it when you feel inspired, and when you don’t, let it rot into internet flotsam and jetsam. It doesn’t need to be a thing! You can even write about Joe Millionaire. I promise I’ll read, or at least skim.

  8. Sam,

    I find that sometimes are questions are TOO simple…

    If the probability was of rolling two dice, and getting, for example, a sum that was odd and less than 3…

    Jonathan

  9. I’m trying to find whether it is possible to represent independent events by venn-diagrams. I have not succeeded so far but would like to receive help from you if you
    know how to do it. Thanks.

  10. Thanks for the post. I came across a similar problem when teaching independence recently. It seems students understand when you have two unrelated events happening a different times (drawing a card and rolling a dice). But when the two events happen simultaneously they run into problems.

    The way I ended up explaining using the dice as an example was to say:
    Look at the numbers 123 456 – Notice half the numbers are less than 4 and half are greater. BUT there are not an equal number odd numbers in both the halves. There are two (1 and 3) that are less than 4; and only one (5) 4 or greater. So IF you know the number rolled is greater than or less than four it provides USEFUL extra information which influence your conditional probability.

    I put all these ideas in a worksheet on my blog if you are interested.
    http://eduflection.blog.com/2014/02/26/independence-and-dependence-of-simultaneous-events/

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