Math Concepts

**Probability considerations and getting a fair grade on a short
quiz.** Download the spreadsheet here.

How much does chance determine whether a student gets a discouraging
score on a quiz? The simplest case may be spelling because teachers
generally do not give partial points. "Tomato" is correct, but "tomatoe"
is wrong. Assume that the students in a class have studied 600 spelling
words over the course of one year, and for the final exam the teacher
tests only a sample of those 600. If a certain student only knows 50% of
the words and the teacher only puts one spelling word on the test, then
that student has a 50-50 chance of getting an A, and a 50-50 chance of
getting an F. The student has no chance of getting a fair grade. If
another student knows none of the words at all, then it doesn't matter how
many spelling words the teacher puts on the final because that student
does not know any of them anyway. If a third student knows absolutely all
of the spelling words, then no matter how many words the teacher tests,
that student will still get 100%.

But what would happen if a student happened to know 85% of the vocabulary
words. If that student were very unlucky, the teacher could make a random
pick of 6 words to put on the test, and all 6 of them would be words that
the student didn't know. If that student were pretty lucky, then the
teacher might pull out 6 words that the student knew, so the grade for the
test would be 100%. The student deserves a B, but he or she might get an A
and might get an F (among other possible grades).

Probability theory can be used to compute the likelihood that the student would get 100%. It turns out to be about 38%. And the probability of getting a 0 turns out to be 0.001%. However, there are lots more ways to get a grade lower than 60%. There is about a 40% chance that the student will get 83% (which is close enough to the deserved grade of 85% that it won't distort the student's transcript very much). So the possibility that the student gets a grade below C will be around 22%.

If the student actually knows 70% of the vocabulary items, then on a
6-item test the student will have about a 12% chance of getting an A, a
30% chance of getting a B, about a 22% chance of getting 67% (which is
not too much worse than the deserved 70%).

If a student only knows half of the spelling words, there is still almost a 2% chance of getting a 100%—or a 0% grade, about a 33% chance of getting a 50% score, and around a 13% chance of getting 83%, 66%, 33%, or 17%.

The attached spreadsheet is initially set up for 85% and 15%, but you can
change those figures to see how the probabilities change.

Here is the rationale for the spreadsheet. The probability that a single
thing will happen might be 1/2. For instance, if one flips a coin, there
is a 1 to 2 chance that it will come up heads, and a 1 to 2 chance that it
will come up tails. If one flips two coins, then the chances are:

H T H T

H H T T

So there is one chance in 4 of getting two heads. That result can be computed by multiplying 1/2 by 1/2.

If there were three coins, then the chances would be

1 2 3 4 5 6 7 8H H H H T T T T

H H T T H H T T

H T H T H T H T

The probability of getting all heads is equal to 1/2 * 1/2 * 1/2 = 1/8, and that result is also indicated in the table above.

Skipping down to a situation in which there are six questions, for any individual student the selected balls (or questions) would have to look like one of the following:

0 0 0 0 0 0 0 0... 5 5 5 5
5 5 5 5 6 6 6 6 6

1 2 3 4 5 6 7 8... 2
3 4 5 6 7 8 9 0 1 2 3 4

H H H H H H H H... T
T T T T T T T T T T T T

H H H H H H H H... T
T T T T T T T T T T T T

H H H H H H H H... H
H H H H T T T T T T T T

H H H H T T T T... H
T T T T H H H H T T T T

H H T T H H T T... T
H H T T H H T T H H T T

H T H T H T H T... T
H T H T H T H T H T H T

In general, the chance of having several things happen together is found by multiplying their individual probabilities. So the chances of throwing three heads would be 1/2 * 1/2 * 1/2 = 1/8, and the chance of throwing the 52nd group of six coins would be 1/2 *1/2 *1/2 *1/2 *1/2 *1/2 = 1/64. There are 64 different combinations that one could get by throwing 6 coins, so in 64 throws the probability of getting all of them should be 1/64 * 64 = 1. All of the chances have been accounted for.

In the case of a large container of black and white balls, if there were 85% black balls and 15% white balls, then experience indicates that it would be easier to draw all black balls, and not very easy to draw all white balls. The calculation is basically the same except that a series of six draws from the large container that would be analogous to the 52nd series of coin throws above would look like this:

5

2

W

W

B

B

W

W

and since there is an 85% chance of drawing a black ball and only a 15% chance of drawing a white ball, the probability of obtaining this collection would be:

85% * 85% *15% *15% * 85% * 85% = 01.17451406%

Since in this collection of "balls" there are four spelling words that
the student taking the test knows and two that the student does not know,
the student's grade would be 66%. The student is actually a B-level
student, but this test result would categorize him or her as a D student.
So the sample size is too small to give reliable results. Increasing the
sample size would improve the "grain" of the test scores by giving more
than 5 possible scores. The student might get one of the other 63 test
configurations, but luck plays too big a factor on the grade the student
would get.

When the probabilities for getting each of the 64 columns on the basis of
the ratio between black and white balls are all calculated and added
together, the sum of probabilities again equals 1, so it is clear that all
possibilities have been included and nothing has been counted twice.

The columns in the spreadsheet that are headed by drawings of dice list the probability that each question will be known or that it will be unknown. The probability that an instructor will make a collection of spelling words a certain number of which are among the 85% of known words and the remainder from among the 15% of unknown words is calculated by multiplying the individual probabilities given in that column. The answer is given at the bottom.