Sunday, February 2, 2014

Keep It Cloud It Junk It (Math Version)

A few weeks ago, Mrs. Parker demonstrated a reading strategy for us.  I've been thinking about a way to change it into a math strategy and I think I've come up with it.

Her strategy was a focus question, then ideas would be set into 3 categories, 1 lead to group discussion, 1 was certain, and the other was extraneous information.  So Cloud It, Keep It, Junk It.  I think this might be applied to word problems like so: Solve It, Use It, Junk It.

Word problems typically have details about the problem and extra information in them just like reading a passage for a purpose.

So for example, look at the following question:

6.  You can win $10 if you can successfully do one of the following given one shot:

I Roll a 6 on a single number cube
II Roll a sum of seven on two number cubes
III Roll a number greater than 4 on a single number cube

Which of these three choices should you have the best chance of getting the $10?
(note, I am not making this offer)

F. I only
G. II and III
H. III only
J. I, II, and III

There's a lot of information here.  Let's split up these things into Solve It, Use It, Junk It.

Solve It                           Use It                        Junk It
P(6)                                best                           $10
P(sum 7)                         number cube              win
P(5 & 6)                         one shot
                                       sum
                                       greater than
Remember that P(6) is shorthand for the probability of getting a 6.  As you can see 3 things are in the solve it category.

The $10 is extra information and doesn't help you solve the problem, so is junked.

Then in Use It, we see that we want to compare the probabilities from the word "best", that they all involve number cubes because no other probability device was used, and that they are not multiple events (one shot).  We also see the word sum (which means add) and greater than.  These are details that modify the probability needed.

After that we solve it by creating ratios.  P(6) = 1/6.  P(sum 7) = 6/36 and P(5 & 6) = 1/3.

Might need to think about this some more.




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