Probability of true/false exam

In summary, the probability of guessing the correct answer on a true/false exam is 50%, or 0.5. The number of questions on the exam affects this probability, as it decreases with more questions. However, the probability remains at 50% as long as the exam is fair. This probability cannot be improved by using a specific guessing strategy and is higher compared to other types of exams.
  • #1
Gridvvk
56
1

Homework Statement


Information in condensed form:

=> Student takes 20 questions true or false exam.
=> He knows answer to 10 questions.
=> He guesses on 10 questions.
=> Professor randomly picks two questions out of 20, and finds student answered them both correctly.

Task: Find probability student knew answer to at least one of the two questions.

Homework Equations


Bayes' Rule and conditional probability definition.


The Attempt at a Solution




My work. I found P(K|C) = 2/3 via Bayes' rule, where K: student knows answer, C: he got it correct. However I'm not sure how to tackle "at least 1" part of problem. My thoughts were pick c1, c2 element of C and find P(K|(c1 or c2)), but this seems like a difficult calculation.
 
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  • #2
Hello,

Could you please explain how you found P(K|C) ?
 
  • #3
Sure.

I defined events as follows
K: Student knows answer
G: Student guesses answer
C: Student gets question correct

P(K) = P(G) = 1/2
Since student knows 1/2 questions and guesses on rest.

P(C|K) = 1, since if student knows answer he is certain to get it correct
P(C|G) = 1/2, half-chance guessing true or false

P(C) = P(C|K) * P(K) + P(C|G) * P(G) = 3/4

P(K|C) = P(C|K) * P(K) / P(C) = (1 * 1/2) / (3/4) = 2/3
 
  • #4
Isn't it just a binomial distribution problem now with N=2 and p=2/3?
 
  • #5
vela said:
Isn't it just a binomial distribution problem now with N=2 and p=2/3?

I thought the binomial distribution only applies when the trials are independent. Wouldn't the probability of getting the second question correctly depend on whether or not the first question was answered correctly or not?
 
  • #6
Yeah, I think you're right.
 
  • #7
Gridvvk said:
Sure.

I defined events as follows
K: Student knows answer
G: Student guesses answer
C: Student gets question correct

P(K) = P(G) = 1/2
Since student knows 1/2 questions and guesses on rest.
That isn't correct, for a couple of reasons. One reason is that the student only answered 20 questions. You have to account for the fact that the universe is finite. The other is that the instructor randomly picks two questions out of the twenty to examine. You need to account for this as well.
 
  • #8
D H said:
That isn't correct, for a couple of reasons. One reason is that the student only answered 20 questions. You have to account for the fact that the universe is finite. The other is that the instructor randomly picks two questions out of the twenty to examine. You need to account for this as well.

I based my formulation on the basis for one particular question out of 20, and hoped I could apply it to two questions. It may be easier to define events for two questions, but I wasn't sure on how to do so in a manner that would allow me to calculate the probabilities.

For instance, if I define an event A such that the student knew the answer to both questions. I'm not sure how to calculate this probability because for one question it would 3/4, but for two questions this probability would decrease.
 
  • #9
Hint: Look at what the instructor drew. He might have picked two questions for which the student knew the answer, two questions where the student was clueless and guessed, or one of each. These are mutually exclusive.

What are the probabilities of these outcomes?
For each of those outcomes, what is the probability that the student got both answers correct?
 
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  • #10
D H said:
Hint: Look at what the instructor drew. He might have picked two questions for which the student knew the answer, two questions where the student was clueless and guessed, or one of each. These are mutually exclusive.

What are the probabilities of these outcomes?
For each of those outcomes, what is the probability that the student got both answers correct?

Thanks a lot, I think I figured it out.

Let Case 1: Student knew answer to both questions
P(Case 1) = (10 C 2) / (20 C 2) = 9/38
P(both correct | case 1) = 1

Let Case 2: Student guessed both questions
P(Case 2) = (10 C 2) / (20 C 2) = 9/38
P(both correct| case 2) = (1/2)^2 = 1/4

Let Case 3: Student knew one question and guessed on other
P(Case 3) = 1 - P(Case 1) - P(Case 2) = 20/38 = 10/19
OR P(Case 3) = [(10 C 1)(10 C 1)] / (20 C 2) = 10/19
P(both correct| case 3) = 1 * 1/2 = 1/2

we want:
P((case 1 or case 3) | both correct) =
[P(both correct | (case 1 or case 3)) * P(case 1 or case 3)] / (P(both correct))

Since the cases are mutually exclusive P(case 1 or case 3) = P(case 1) + P(case 3) = 29/38

P(both correct | (case 1 or case 3)) =
P(both correct and (case 1 or case 3)) * P(case 1 or case 3) =
P((both correct and case 1) or (both correct and case 3)) / P(case 1 or case 3) =
[P(both correct and case 1) + P(both correct and case 3)] / P(case 1 or case 3) =
[P(both correct|case 1) * P(case 1) + P(both correct|case 3) * P(case 3)] / P(case 1 or case 3)
= [1 * 9/38 + 1/2 * 20/38] / [29/38] = 19/29

P(both correct) = P(both correct|case 1) * P(case 1) + P(both correct|case 2) * P(case 2) + P(both correct| case 3) * P(case 3) = (1 * 9/38) + (1/4 * 9/38) + (1/2 * 20/38) = 85/152

so...back to what we wanted
P((case 1 or case 3) | both correct) =
[P(both correct | (case 1 or case 3)) * P(case 1 or case 3)] / (P(both correct)) =
(19/29 * 29/38) / (85/152) = 76/85.

About 89.4%
 
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  • #12
Gridvvk said:
Thanks a lot, I think I figured it out.

Let Case 1: Student knew answer to both questions
P(Case 1) = (10 C 2) / (20 C 2) = 9/38
P(both correct | case 1) = 1

Let Case 2: Student guessed both questions
P(Case 2) = (10 C 2) / (20 C 2) = 9/38
P(both correct| case 2) = (1/2)^2 = 1/4

Let Case 3: Student knew one question and guessed on other
P(Case 3) = 1 - P(Case 1) - P(Case 2) = 20/38 = 10/19
OR P(Case 3) = [(10 C 1)(10 C 1)] / (20 C 2) = 10/19
P(both correct| case 3) = 1 * 1/2 = 1/2

we want:
P((case 1 or case 3) | both correct) =
[P(both correct | (case 1 or case 3)) * P(case 1 or case 3)] / (P(both correct))

Since the cases are mutually exclusive P(case 1 or case 3) = P(case 1) + P(case 3) = 29/38

P(both correct | (case 1 or case 3)) =
P(both correct and (case 1 or case 3)) * P(case 1 or case 3) =
P((both correct and case 1) or (both correct and case 3)) * P(case 1 or case 3) =
[P(both correct and case 1) + P(both correct and case 3)] * P(case 1 or case 3) =
[P(both correct|case 1) * P(case 1) + P(both correct|case 3) * P(case 3)] * P(case 1 or case 3)
= [1 * 9/38 + 1/2 * 20/38] * [29/38] = 29/76

P(both correct) = P(both correct|case 1) * P(case 1) + P(both correct|case 2) * P(case 2) + P(both correct| case 3) * P(case 3) = (1 * 9/38) + (1/4 * 9/38) + (1/2 * 20/38) = 85/152

so...back to what we wanted
P((case 1 or case 3) | both correct) =
[P(both correct | (case 1 or case 3)) * P(case 1 or case 3)] / (P(both correct)) =
(29/76) / (85/152) = 58/85.

About 68.2%

Well, I get an answer of 76/85 ≈ 0.894.

Here is what I did: for k guessed correct, the population consists of three classes: (1) 10 known & correct; (2) k guessed and correct; (3) 10-k guessed wrong. The probability of k guessed correct is the binomial probability Pk = C(10,k)/2^10, where C(a,b) denotes the binomial coefficient "a choose b".

Denote by P(u,v,w) the probability that the prof. chooses u from class 1, v from class 2 and w from class 3. So, conditioned on k guessed OK, the probability that the prof chooses 2 correct is P(2,0,0)+P(1,1,0)+P(0,2,0). These are 3-class hypergeometric probabilities: that is, P(u,v,w) = C(10,u)*C(k,v)*C(10-k,w)/C(20/2) [all conditioned on k guessed OK]. Thus, we have P(2,0,0) = (10/20)(9/19), P(1,1,0) = 2(10/20)(k/19), and P(0,2,0) = (k/20)((k-1)19)= k(k-1)/(20*19).

The probability that the prof chooses 2 correct and the student knows at least one is P(2,0,0) + P(1,1,0).

So: P{2 correct} = sum_{k=0}^10 C(10/k)/2^10 * [P(2,0,0)+P(1,1,0)+P(0,2,0)] = 85/152.
P{2 correct and >=1 known} = sum_{k=0}^10 C(10,k)/2^10 *[P(2,0,0)+P(1,1,0)] = 1/2.
Thus, the conditional probability P{>=1 known|2 OK} = (1/2)/(85/152) = 76/85.
 
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  • #13
Ray Vickson said:
Well, I get an answer of 76/85 ≈ 0.894.

Here is what I did: for k guessed correct, the population consists of three classes: (1) 10 known & correct; (2) k guessed and correct; (3) 10-k guessed wrong. The probability of k guessed correct is the binomial probability Pk = C(10,k)/2^10, where C(a,b) denotes the binomial coefficient "a choose b".

Denote by P(u,v,w) the probability that the prof. chooses u from class 1, v from class 2 and w from class 3. So, conditioned on k guessed OK, the probability that the prof chooses 2 correct is P(2,0,0)+P(1,1,0)+P(0,2,0). These are 3-class hypergeometric probabilities: that is, P(u,v,w) = C(10,u)*C(k,v)*C(10-k,w)/C(20/2) [all conditioned on k guessed OK]. Thus, we have P(2,0,0) = (10/20)(9/19), P(1,1,0) = 2(10/20)(k/19), and P(0,2,0) = (k/20)((k-1)19)= k(k-1)/(20*19).

The probability that the prof chooses 2 correct and the student knows at least one is P(2,0,0) + P(1,1,0).

So: P{2 correct} = sum_{k=0}^10 C(10/k)/2^10 * [P(2,0,0)+P(1,1,0)+P(0,2,0)] = 85/152.
P{2 correct and >=1 known} = sum_{k=0}^10 C(10,k)/2^10 *[P(2,0,0)+P(1,1,0)] = 1/2.
Thus, the conditional probability P{>=1 known|2 OK} = (1/2)/(85/152) = 76/85.

Thanks for the catch. I got that answer first as well, but thought I made an arithmetic mistake, so I got the new one, but looking back 76/85 was indeed correct. Seems the methods are similar, but your terminology and formulation is better, because you can generalize this to variations of the problem. Thanks for the help.
 
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  • #14
Ray Vickson said:
Well, I get an answer of 76/85 ≈ 0.894.
That was the answer posted when I wrote "very good". I would not have written that had Gridvvk posted an answer of 58/85. Apparently Gridvvk wavered on the result for a while.
 

1. What is the probability of guessing the correct answer on a true/false exam?

The probability of guessing the correct answer on a true/false exam is 50%, or 0.5. This assumes that the exam is fair and that the true and false options are equally likely to be correct.

2. Does the number of questions on the exam affect the probability of guessing the correct answer?

Yes, the number of questions on the exam does affect the probability of guessing the correct answer. As the number of questions increases, the probability of guessing all of them correctly by chance decreases. For example, on a 10-question true/false exam, the probability of guessing all 10 correctly is 0.5 raised to the power of 10, or 0.0009765625.

3. How does the probability of guessing the correct answer change if the exam has more true or false questions?

If the exam has more true or false questions, the probability of guessing the correct answer will still be 50%, as long as the exam is fair. However, the chances of guessing all of the questions correctly by chance will decrease as the number of questions increases.

4. Can the probability of guessing the correct answer on a true/false exam be improved by using a specific guessing strategy?

No, the probability of guessing the correct answer on a true/false exam cannot be improved by using a specific guessing strategy. This is because the probability is already 50% and is not affected by any strategy. However, it is still important to carefully read and understand the questions in order to eliminate any obvious incorrect answers.

5. How does the probability of guessing the correct answer on a true/false exam compare to other types of exams?

The probability of guessing the correct answer on a true/false exam is higher than other types of exams, such as multiple choice or essay exams. This is because there are only two options to choose from, making it more likely to guess correctly. However, this does not mean that true/false exams are easier, as they still require a good understanding of the material in order to select the correct answer.

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