Probability of passing PE with random guessing

So, I was curious what the probability of someone passing the PE exam based on completely random guessing (because it seemed feasible), so, I did some math. The probabilities are based on an assumed passing score of 70% (56/80), using the binomial cumulative probability... that is, for randomly guessing all 80 questions, what's the probability of guessing at least 56 correct? Thus, calculate the binomial cdf for (n=80, p=0.25, and x=56 + x=57 + ...x=80)

For all 80 questions, the probability of guessing at least 56 correctly turns out to be 3.64 x 10^-17.  That is, 1 in 27,500,000,000,000,000 test takers will be likely to pass the exam with random guessing.

However, if a test taker is 100% confident that he worked out half the questions correctly, and wanted to simply guess on the other 40 questions, he would only need to guess 16 problems correctly out of the remaining 40. The probability is calculated similarly, with a binomial cdf of (n=40, p=0.25, and x=16 + x=17....x=40). This turns out to be a probability of 0.026. That is, 1 out of about 50 test takers who definitely answer 40 questions correctly can expect to pass the exam by guessing on the remaining questions.

Here's the breakdown:

0 definitely correct, Guess 56/80 correctly to pass : 1 in 27,500,000,000,000,000  will pass(1 out of 27 Quintillion)

10 correct, Guess 46/70: 1 in 1,170,000,000,000 will pass (1 out of a trillion)

20 correct, Guess 36/60: 1 in 103,000,000 will pass

30 correct, Guess 26/50: 1 in 26,300 will pass

40 correct, Guess 16/40: 1 in 50 will pass.

50 correct, Guess 6/30, 4 in 5 will pass.

60 correct ... pretty sure you passed!

My only comment is that calculation needs to be broken down into the following when guessing:

1. Narrowed answers down to 2 choices (eliminated 2 that were obviously incorrect) and made an educated guess (50% chance of making the correct choice.

2.  Random Guess (25%  chance you made the correct choice)

Also, I read on another board that NCEES equalizes the answers so that there is an equal distribution of answers (20-A's, 20-B's, 20-C's, and 20-D's).  That way, there is no statistical advantage to guessing a specific answer (for example all C's).  

When I had to random guess the last couple of questions, I did a quick count of the answers and then guessed so that my answers had close to an equal distribution.

After the test, I broke it down as follows:

1.  60 questions I got an answer that matched one of the answers provided.  I went ahead and assumed that I got 80% of these correct.  60 x .80 = 48 correct

2.  10 questions I had the answer down to 2 possibly answers.  I assumed that I got 50% of these correct. 10 x .50 = 5 correct

3.  10 questions I got an answer that did not match one of the answers provided or I simply ran out of time.  I guessed on these based on the equal distribution theory. I assumed I got 25% of these correct. 10 x .25 = 2.5 correct.  Rounding up that is 3 correct.

48+5+3 = 56 correct.  

Then I slept on and thought the following:

1.  64 questions I got an answer that matched one of the answers provided.  I went ahead and assumed that I got 80% of these correct.  64 x .80 = 51 correct

2.  4 questions I had the answer down to 2 possibly answers.  I assumed that I got 50% of these correct. 4 x .50 = 2 correct

3.  16 questions I got an answer that did not match one of the answers provided or I simply ran out of time.  I guessed on these based on the equal distribution theory. I assumed I got 25% of these correct. 16 x .25 = 4 correct.

51+2+4= 57 correct

Then I thought maybe I only got 75% instead of 80% correct, convinced myself I failed, slept on it, then woke up and thought maybe I got 85% correct, slept on it, and so on and so on.  Finally I realized I have no freaking clue how I did, just that the exam sucked and I hoped to never have to take it again, which thankfully I did not have too. 

ruggercsc said:

Also, I read on another board that NCEES equalizes the answers so that there is an equal distribution of answers (20-A's, 20-B's, 20-C's, and 20-D's).  That way, there is no statistical advantage to guessing a specific answer (for example all C's).  

Click to expand...

Interesting, so, in the unlikely event that someone is 100% sure of answering 50 questions correctly, they could potentially guarantee passing by guessing the same answer for the remaining questions... assuming they're smart, and guess the answer that was least frequently already used.

That sounds correct, if the equal distribution theory holds true.  

Typical scenarios would be as follows:

• A's - 10/B's - 10/C's- 15/D's -15:  Answer all B's for the next 30, then 10 would be correct and you would have 60/80.
  
• A's - 11/B's - 13/C's- 13/D's -13:  Answer all A's for the next 30, then 9 would be correct and you would have 59/80.
  
• A's - 0/B's - 10/C's- 20/D's -20:  Answer all A's for the next 30, then 20 would be correct and you would have 70/80.
  
• A's - 1/B's - 19/C's- 15/D's -15:  Answer all A's for the next 30, then 19 would be correct and you would have 69/80.
  
• A's - 13/B'- 12/C's - 12/D's -13:  Answer all C's for the next 30, then 8 would be correct and you would have 58/80.
  

As long as 58 is the cut score, I believe this you would pass if you knew you got 50 correct and had to guess with the least frequently used answer for the rest.  Maybe someone can see if there is a scenario where this does not work (assuming the equal distribution of answers holds true). 

It probably doesn't matter. As I showed in the original post, if you know you have 50 correct, the probability of passing by completely random guessing is already nearly 80%.

My takeaway from this... if you can be really confident with 50-or-so of the questions on the exam, you have a really high chance of passing by just guessing on the rest. Even if you can be very confident on 40 of the questions, the probability of passing is not prohibitively low. 

If my calculations are correct (and I'm not sure that they are), the probability of guessing 16/40 increases to 6% (1 in 16) if you can narrow down 10 of the answers to a 50/50 guess. That was calculated assuming the 56 cut-off, and using all possible combinations of the 10 question with 50% probability and 30 questions with 25% probability that will net a passing score(e.g. 0/10 AND 16/30, or 1/10 AND 15/30 ... or 10/10 AND 6/30)

The way I looked at it before I took my exam was:

Let X = the minimum number of questions you need to nail

[ X + .25(80 - X) ] / 80 = .70,  X = 48

(Assuming you need a 70% to pass and assuming you will get 25% of the questions you don't nail)

ptatohed said:

The way I looked at it before I took my exam was:

Let X = the minimum number of questions you need to nail

[X + .25(80 - X) ] / 80 = .70,  X = 48

(Assuming you need a 70% to pass and assuming you will get 25% of the questions you don't nail)

Click to expand...

Interesting in that my first analysis assumes one had to get at least 48 answers that you knew the answers too before you started guessing.  Your method is more simpler and I tend to over analyze.      

ptatohed said:

Let X = the minimum number of questions you need to nail

[ X + .25(80 - X) ] / 80 = .70,  X = 48

(Assuming you need a 70% to pass and assuming you will get 25% of the questions you don't nail)

Click to expand...

While I like this simple calculation, the assumption of getting 25% of guessed answers correct is a big one. For the 32 remaining questions, considering all possible combinations of successful guesses at 25% probability, and unsuccessful guesses at 75% probability, the probability of guessing exactly 8 correct is somewhat low: 16%. However, the probability of guessing at least 8 correct is not too bad, at 57%. So, you're slightly more likely to pass than fail if you nail 48 questions.

48 indeed seems to be the magical cut-off. If you nail 47 questions, the probability of guessing at least 9 correct out of the remaining 33 questions turns out to be 45%, thus, you'd be slightly more likely to fail in that scenario.

So, if you're a "D" student (i.e. you're really comfortable with only 60% of the material on the exam), it seems you're still pretty likely to pass the exam by random guessing alone. I think I'm OK with that...D's get Degrees, or something like that. Anything less than nailing 60% of the questions, and the probability of passing based of random guessing goes downhill very quickly. Of course, if you're a "C" student, or better, than you don't need to worry about any of this.

no guessing. 

Engineered guessing works for optimal results.

From my experience I saw similar questions that I have to guess. I choose different answers (even feel they are opposite) to at least get one right.

Seems folks have a lot of time on their hands

Similar to the equation I placed above ( [ X + .25(80 - X) ] / 80 = .70 ), when I took each of my exams (CA exams and NCEES exam), this is what I did to "know" if I passed or not.

While taking the exam, I put a little mark (a pencil dot) next to each question number (on the scantron) that I felt "super, duper good" about getting correct.  If I felt just "kinda good", or I got it to a 50/50, I did not mark it.  I gave myself full credit for the "super, duper feel goods" and 25% for the others.  The hope was that I was pretty accurate because while I'm sure I still missed a few "super, duper feel goods", by only giving myself 25% of the "feel kinda goods" or the 50/50s, it would average out.   

I distinctly remember for the NCEES 8hr getting 27 AM and 28 PM "super, duper feel goods".

[ 27 + .25 (40 - 27) + 28 + .25 (40 - 28) ] / 80 = 61.25/80 (say 61/80) = 76.3%. 

I assumed I passed, and I did.  Of course I'll never know how close my actual score was to 76.3%.  But it gave me piece of mind waiting for results.    

anarchonobody said:

0 definitely correct, Guess 56/80 correctly to pass : 1 in 27,500,000,000,000,000  will pass(1 out of 27 Quintillion)

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You better change winning the powerball or mega millions than get 56 rights