Probability of Guesser Scoring 100% on Midvale School Tests

[Philip Wong]

Homework Statement

Midvale School for the Gifted has two types of students: Guessers and Swots. All
Midvale tests consist of sets of questions with yes/no answers. Guessers will simply
answer yes or no to each question as the mood takes them, so they have probability
0.5 of getting each question correct. Swots, on the other hand, will certainly get
every question correct with probability 1. Dr Ramble Fluster is the Chief Examiner
at Midvale. She worries about her test results. Clearly, all Swots will get 100% on
every test. However, it is possible that a Guesser might get 100% just by chance.
Ramble's job is to decide how many questions to set in each test to be fairly certain
that she will be able to give the correct grade (G or S) to each student.

(b) What is the probability that a Guesser scores 100% on a test with n questions?

Homework Equations

Binomial function

The Attempt at a Solution

X~Bin(n,0.5)

P(X=x)= nCx * p^x * (1-p)^(n-x)

P(X=100) = nC100 * 0.5^100 * 0.5^(n-100)

here is where I got stuck. Since the question is asking us to solve the probability equation (P(X=x)), but since n is an unknown I don't know how to go further into solving this equation, because (I presume) there will be 2 unknown parameters: P and n.

can someone point me where I get wrong, or give me an idea on how to solve this?
thanks

 

[Dick]

nC100?? Think about what that means. You aren't choosing 100 percentage points from n percentage points. Scoring 100% means you got all n questions right. Shouldn't the probability involve nCn?

 

[Philip Wong]

nC100?? Think about what that means. You aren't choosing 100 percentage points from n percentage points. Scoring 100% means you got all n questions right. Shouldn't the probability involve nCn?

ar! ok, I've just reattempt this question from what I understand in your suggestion. Which doesn't seems to be correct though, need more help thanks.

reattempts are as follows:

fx(x)= p(X=x) = nCx * p^x * (1-p)^n-x

let x = n

p(X=n) = nCn * p ^ n * (1-p)^n-n
= 1 * 0.5^n * 0.5^0
= 0.5^n

 

[Dick]

ar! ok, I've just reattempt this question from what I understand in your suggestion. Which doesn't seems to be correct though, need more help thanks.

reattempts are as follows:

fx(x)= p(X=x) = nCx * p^x * (1-p)^n-x

let x = n

p(X=n) = nCn * p ^ n * (1-p)^n-n
= 1 * 0.5^n * 0.5^0
= 0.5^n

That seems ok to me. E.g. if there are two questions P=(1/4)=(0.5)^2. You have to get the first question right AND the second question right - both with probability 1/2.

 

[Philip Wong]

That seems ok to me. E.g. if there are two questions P=(1/4)=(0.5)^2. You have to get the first question right AND the second question right - both with probability 1/2.

ah! looks light you are right. because I just read the next part of the question "Ramble wishes to select the number of questions, n, so that there is less than 0.5% chance that a Guesser gets full marks. What is the lowest value of n that will guarantee this?"

so I will just substitute in the numbers and found out what n is.

thanks a lot for your help!

 

[Dick]

You're welcome. You could solve it systematically using a log. But there's nothing wrong with substituting either.