Q&A #3784

Teachers' Lounge Discussion: Tree diagram: Four Children (2B's/2G's any order)

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From: Dave Marain

To: Teacher2Teacher Public Discussion
Date: 2001032421:19:17
Subject: Re: tree diagram

Hi! Tree diagrams provide a mechanical structure but, for an introduction to combinatorial problems, we probably want to help our students develop conceptual understanding BEFORE procedures or formulas. Thus with 2 children (assuming 50-50 chances for a boy or girl), we can throw 2 coins in the air and ask the students to think about and then discuss with their "partner" what that experiment has to do with the gender of children! To promote further concept development, emphasize the qualitative aspect by asking, which is more likely: Both Heads, Both Tails, or 1 of each. Of course the children could act this out 100 times and record the data, but most children will see the point if you jump to an extreme case (a powerful mathematical tool!): What if I throw 100 coins in the air; what is most likely, 100 heads, 100 tails or some of each?!? Again, before you get to the tree diagram "procedure," ask the students to explain to their partners, why 1 head, 1 tail is more likely than both heads. Some youngsters will realize that there are TWO ways to get one of each coin (or one of each gender), whereas there is only ONE way to get both heads or both tails. This is even more apparent when you use DIFFERENT coins, like a penny and a nickel. This way it is easier to distinguish the outcomes: Heads-Penny, Heads-Nickel is the ONLY way to get both heads (order is not an issue at this point), whereas, Heads-Penny, Tails-Nickel or Heads-Nickel, Tails-Penny give the TWO ways for getting one of each kind. Note that this discussion DOES NOT TAKE ORDER INTO ACCOUNT! It assumes that both coins are being toosed SIMULTANEOUSLY! Tree diagrams and similar counting procedures ALWAYS solve the problem by considering the various ORDERS that are possible. Now to the tree method for 3 children: 1st level: B G 2nd level: B G B G 3rd level: B G B G B G B G There is a total of EIGHT paths through the tree. Let the group discover this with 2 children first (2 x 2 = 4), then some will make the leap to 3 children (2 x 2 x 2 = 8). [Note: By middle school, they should have had many experiences with the multiplication principle of counting.] There is only ONE path that leads to ALL boys and ONE path to ALL girls. There are THREE paths that lead to 2 boys and 1 girl (considering all possible orders); similarly, THREE ways to have 2 girls and 1 boy. Thus the probabilities are 3 boys: 1/8 3 girls: 1/8 2 girls, 1 boy: 3/8 2 boys, 1 girl: 3/8 Let me know if this seems reasonable for your students. What would you do differently? What other manipulatives could we use? Dave M

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