Math Question (Probability)

I have to take an essay test for my history class. I was given a study guide with seven possible questions to study, four of which will be on the test. From those four, I must chose one question to answer.

Now, logic tells me that I only need to study four of the seven questions, since I will then be guaranteed to know at least one question on the test, since only three of the seven questions will not be on the test.

My question is, if I were to study only three of the seven study questions, what is my probability of getting at least one of those three on the test? How about if I only study two? Or one? How would I go about calculating this other than trying to figure out every possible combination of questions?


There are 35 possible combinations of 4 questions that can be on the quiz: 7!/(3! * 4!).

Of those 35, any one particular question will be on 20 of those combinations. So if you only choose one question to study, there’ll be a 20/35 or 57.14% chance that it will be on the test.

If you study only 2, at least 1 of those questions will be on 30 of the 35 combinations, or an 85.71% chance.

If you study only 3, at least 1 of those questions will be on 34 of the 35 cominations, or a 97.14% chance.

And, as you surmised, if you only study 4, at least one of them will definately (100%) be on the test.

I’ll take a stab at it.

If you study for one, then there is a 3 out 7 (3/7) chance that the one you prepared for is not on the exam.

If you study for two, there is a (3/7)*(2/6) = 1/7 chance that neither is on the exam.

If you study, there is a (3/7)(2/6)(1/5) = 1/35 chance that none is on the exam.


Imagine a box with 4 red balls (questions on the exam) and 3 white balls (questions not on the exam). If you choose one ball at random, there’s a 3/7 chance it was white. The chance of getting two consecutive whites are 3/7 X 2/6. The chance of three consecutive whites is 3/7 X 2/6 X 1/5.

First, the answer. P[sub]n[/sub] is the probability of being okay on the test if you study n problems.

P[sub]4[/sub]=1 (100%)
P[sub]3[/sub]=34/35 (97%)
P[sub]2[/sub]=6/7 (86%)
P[sub]1[/sub]=4/7 (57%)
P[sub]0[/sub]=0 (0%)

You definitely don’t need to enumerate all possibilities. The answer is the same as if you have a bag with four red balls and 3 white balls. If you draw three out, what’s the chance that you get at least one red one. Well, it’s 1 minus the probability that you get all white ones. Example:

P[sub]2[/sub]=1-(3/7)(2/6) since the probability of drawing white ones on both shots is 3/72/6 = 1/7.

(Looks like a simulpost…)

BTW, I did my calcs by counting the combos, since there were few enough to do this quickly.

If you want the math, I suppose 6!/(3! * 3!) will tell you how many times a particular question will be amongst the combinations. Beyond that, I don’t have the brain cells functioning this late on a Sunday to work that out. :D:D

I see several other answers have arrived while I was writing this, but I’m explaining hypergeometrics in general, rather than the special case “0 choice” answers the other posters have given you, so I’ll do it anyway.

I’m going to use nCk as a notation for "combination of n things taken k at a time. The more normal notation is the two numbers one over top of each other in parentheses, but that’s cumbersome for text-based discussions.

nCk = n! / (k! * (n-k)!)

We can arrive at that by realizing that n! / (n-k)! counts the ORDERED arrangement of the objects: n choices for the first slot, (n-1) for the second, etc, down to n-k. Since we want the combinations without regard to order, we divide by k! since each unordered combination will show up in the ordered list in each of k! orders.

Now, let’s say we choose to split our n up into two categories, like questions you chose to study and questions you didn’t. Let’s designate the number of items in the “special” category as m. Then, the number of combinations representing exactly p of the “special” ones is the number of ways of choosing p from the m special items times the number of ways of choosing k - p from the remaining n - m items: mCp * (n-m)C(k-p).

So, that, divided by nCk will be the probability of getting exactly p of the m “special” items:

= mCp * (n-m)C(k-p) / nCk

In fact, this is called a “hypergeometric” distribution, and you should learn about it in a math class sometime.

Now, apply to your problem. There are 7C4 = 35 combinations which could appear on the test. If you study 3 of them, we have 3 “special” ones, and you are screwed if none of them turn up. The probability of that is

3C0 * 4C4 / 7C4 = 1/35

If you study 2 of them:

2C0 * 5C4 / 7C4 = 5/35

If you study 1 of them:

1C0 * 6C4 / 7C4 = 15/35

Now go study your damned history!

Thanks everybody. This board is a wonderful procrastination tool. Nothing beats learning math when you’re supposed to be learning history.

Hmmm… 97.14% you say? Those are good odds. I’m a gambling man…

Congratulations, Caliban. For the first time in SDMB history, the first two answers to a thread about probability have agreed. Most probability threads turn into hideous monsters of nitpicking, fine points, and semantics… Anyone else remember the Monty Hall problem? shudder

By the way,

I think that this is the funniest thing I’ve read today.