kitchen table math, the sequel: probability redux

## Monday, August 30, 2010

### probability redux

16. If j is chosen at random from the set {4, 5, 6} and k is chosen at random from the set {10, 11, 12}, what is the probability that the product of j and k is divisible by 5?

SAT practice test
I missed this one.

Catherine Johnson said...

I missed only two on two of the 3 math sections today!

If I did that on a timed test, my score would be in the mid 700s.

One of the questions I missed was incredibly easy ... it was a 'careless error' miss that I probably would have gotten right under pressure.

Didn't know how to do this one - and in fact didn't even notice I didn't know how to do it until now.

Anonymous said...

There are 9 possible pairings. The ones divisible by 5 are
5 x anything (3)
4,6 x 10 (another 2)
Tricky point: don't count 5x10 twice!

Probability: 5/9

Allison said...

the product jk is divisible by 5 if either j or k is divisible by 5.

What is the sample space?

The sample space is the set of all products formed from j and k.

How many elements are there in the sample space?
3*3 = 9, because you can choose any element of j and any element of k.

How many elements in the sample space correspond to the event?

Elements in the event are when either j or k is divisible by 5. Notice the or: how can you rewrite the or has a set of mutually exclusive events?

j divis + k divis - j and k divis.

does that help?

Glen said...

Three first nums, three second, 3x3 is nine pairs. All you have to do, if you have memorized your 12x12 multiplication tables, is count up to a maximum of nine on your fingers as you go through the list:

4x10 yes
4x11 no
4x12 no

5x10 yes
5x11 yes
5x12 yes

6x10 yes
6x11 no
6x12 no

5 yeses out of 9, or 5/9. Done. Next.

This is not how I would do a bigger or more complex problem, but this is a small problem on a timed test. Rote memory and fingers--the original digital accumulators.

Of course I just put up three more fingers at once when I got to the three 5x? combos in a row instead of multiplying each one, but that hardly mattered. On a timed test, any further analysis of a problem this small (beyond rote recall and counting on fingers) would just come at the expense of subsequent problems.

Catherine Johnson said...

Thanks so much --- all of you --- this is incredibly helpful.

Glen - that's what I was wondering: is it best just to write out the possibilities ---- ?

Catherine Johnson said...

j divis + k divis - j and k divis

I'm still mulling this -- I'll have to write it out for myself.

I need to re-do the Dolciani chapter -- actually, I think I'll work my way through the Arlington Algebra sequence first & study the pages lsquared directed me to.

Glen said...

It's best to avoid writing the list if you can. Even a short list like this takes too much time and, worse, is error-prone, if you are writing under time pressure.

It's best if you can get comfortable systematically walking down the list of all combinations mentally, for various types of situations, because these problems are so common on tests. Having mental models such as tree structures to guide you is essential so you are confident that you will list each possibility once, and only once, with no invalid items.

I'm not sure what to suggest you read to find these techniques. It's usually covered in discrete math books. It has been so long that I don't remember, but I think I must have learned most of it from programming algorithm books decades ago. Exhaustively walking through all possibilities, accumulating items, or at least incrementing a counter, each time one matches your criteria, is extremely common in programming.

Most discrete math and programming books are overkill for you, though, so maybe someone could suggest a better resource for the basics likely to show up on these sorts of tests.

Anonymous said...

P(j=6) + P(k=12) - P( (j,k) = (6,12)) =
1/3 +1/3 - 1/9 =5/9.

"what allison said", in other words.

vlorbik (MEdZ).

Anonymous said...

arg. i did "divisible by 3"
in my writeup. i *mean*
P(j=5) + P(k=10) - P( (j,k)=(5,10) ).

of course.
v.

Michael Weiss said...

I visualize a 3 x 3 grid -- a multiplication table with {4, 5, 6} along the top and {10, 11, 12} going down the side. There are 9 products you can form. Everything in the "5" column and everything in the "10" column is divisible by 5. Count the spaces: there are five of them.

I solved this a completely different way.

P(product is divisible by 5) = 1 - P(product is not divisible by 5)

P(product is not divisible by 5) = P(5 is not chosen from {4,5,6} AND 10 is not chosen from {10,11,12})

= P(5 is not chosen from {4,5,6}) * P(10 is not chosen from {10,11,12})

= 2/3 * 2/3

= 4/9

So P(product is divisible by 5) = 1 - 4/9 = 5/9

It actually took about 3 seconds to do in my head, but 3+ minutes to write down the steps.