There are 7 choices and 3 right answers, what is the probability that out of 3 guesses I choose 2 right answers?

Question

Probability is a pretty practical subject matter, so I decided to apply it more and more to daily life. So while watching a documentary on extreme sports the other day, a photograph of 7 people were shown and the documentary voiceover guy said that of those 7, 3 died via the extreme sport they par-took in (which I believe was a modified version of sky diving). I immediately paused the documentary, and tried to guess which 3 had died, and I guessed 2 right. Now taking ai-class, and learning probability and all, I got curious as to what the probability was of what I had done. Basically, I wanted to know how likely it was I guessed 2 right just by sheer chance (given the 3 guesses I had and the seven possibilities in total) (to see how well I had "profiled" the 7 sky divers). How would I do this in terms of combinations/permutations?

Accepted Answer

2

There are 7 choose 3 (35) possible groups that you could have made with three guesses. The number of groups of three in which two guesses were correct is 3 choose 2 (3), but for each of those groups there are 4 ways that you could have slotted one wrong guess in (eg, if Adam, Bob and Claire died, and you guessed Adam and Claire correctly, you're third guess could be Dave, Ellen, Frank or Gretchen), so there are 12 ways to correctly guess 2 out of three, out of 35 possible guesses of 3 from 7, so the odds that you would guess two correctly are 12/35 or roughly 34%. The odds that you would guess at least two correctly is 13/35, or ~37%

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answered 17 Sep '11, 15:48

gregmchapman ♦
976●1●7●14

I thought about what you said for a few minutes, and then it made complete sense. The part that was confusing me was the multiplying by 4 part. But now I get it: there are 4 people I can pick as a wrong answer each time I get 2 people right or 2 right answers. Thanks a lot! So since the probability of getting at least 2 right is less than 50% I will take it that I beat the odds ( +1 for my "profiling" skills :] ).

(17 Sep '11, 16:42) burton

Answer 2

4

This is an instance of the "urn problem without replacement" – since every time you select a "candidate" the population decreases by one – and as such it is governed by the hypergeometric distribution. In the terms of your problem, for n "candidates" chosen from an original group of size N (among which m are dead), the probability that k of the "candidates" you chose are dead is given by:

$hypergeometric distribution formula$

Substituting your case's parameters in the above equation and calculating the binomial coefficients you get (I'm switching to Python so my results are more easily reproducible):

from math import factorial

def binomial(n, k):
    return float(factorial(n) // (factorial(k) * factorial(n - k)))

n = 3
N = 7
m = 3
k = 2

X = binomial(m, k) * binomial(N - m, n - k) / binomial(N, n)

print 'X = {0!s}'.format(X)

---

X = 0.34285714285714286

So informally the answer to your question is "just by choosing randomly, you should get it right about once every 3 tries".

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answered 17 Sep '11, 17:29

xperroni ♦
1.4k●2●11●23

edited 17 Sep '11, 17:31

Oh sure, bring mathematical rigor into this :)

(17 Sep '11, 17:36) gregmchapman ♦

The python code definitely helps (I can use it for any other "urn problem without replacement"). The hypergeometric distribution you mentioned, I get conceptually, but will need to look more into mathematically. Thanks tho :) (and lol @gregmchapman).

(18 Sep '11, 14:37) burton

It's not as complicated as it sounds. Essentially what the formula is calculating is the reason:

(All possible "right" outcomes)
-------------------------------
    (All possiple outcomes)

How it arrives at it is a bit tricky, but bear with me.

In the formula above, the binomial coefficient operator $binomial coefficient operator$ gives the number of distinct subsets of size k that can be created from a finite set of size n. By definition, two subsets a and b are distinct if there is at least one element i that is contained in a and not by b, or vice-versa – element order is not a factor.

First we calculate the number of all possible test outcomes that comply to the case parameters. In your scenario we want two guesses right out of three, so we take the number of all possible subsets of size 2 containing only "right" guesses, and multiply by the number of all possible subsets of size 1 (3 - 2) that contain only "wrong" guesses. The we divide the result by the number of all possible outcomes (i.e. all possible three-guesses results).

(19 Sep '11, 07:54) xperroni ♦

Answer 3

I suppose you want two and only two right guesses from the tree picks, so there are 3 possible outcomes (along with their proper probability below):

hit    hit   miss:
3/7 *  2/6 *  3/5
hit    miss   hit:
3/7 *  4/6 *  2/5
miss   hit    hit:
4/7 *  3/6 *  2/5

These possibilities all add up to the total probability that we look for, so the result is:

(3*2*3 + 3*4*2 + 4*3*2)/(7*6*5) = 11/35

So the only two right guesses you will get with probability of 11/35 ~ 0.314

If you allow also the possibility of all the picks to be right guess, you can add as well:

hit   hit   hit:
3/7 * 2/6 * 1/5

for a total of: 12/35 ~ 0.342

There are 7 choices and 3 right answers, what is the probability that out of 3 guesses I choose 2 right answers?

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