Probably some binomial distributions.

Hello. My problem:
Student can answer each question given to him with a probability of 0.4.
A student is asked till he gives a correct answer.
What is the probability that no more than 4 questions will be given.
If we use P(X)=nC​r​*p^x​​*q^​n−x​​ then p=4? q=0? n=? x=? no idea.
On the other hand (0.4)+(0.6*0.4)+(0.6*0.6*0.4)+(0.6*0.6*0.6*0.4)=0.87,
which is approx 87% might look correct but that's also probably wrong since
such calculation can be done only for independent probabilities and ours is
clearly not independent since later questions depend on answer of previous questions.
Anyway I would greatly appreciate any help.

xeonas said:

Hello. My problem:
Student can answer each question given to him with a probability of 0.4.
A student is asked till he gives a correct answer.
What is the probability that no more than 4 questions will be given.
If we use P(X)=nC​r​*p^x​​*q^​n−x​​ then p=4? q=0? n=? x=? no idea.
On the other hand (0.4)+(0.6*0.4)+(0.6*0.6*0.4)+(0.6*0.6*0.6*0.4)=0.87,
which is approx 87% might look correct but that's also probably wrong since
such calculation can be done only for independent probabilities and ours is
clearly not independent since later questions depend on answer of previous questions.
Anyway I would greatly appreciate any help.

Click to expand...

Questions like this are generally approached from the opposite end. That is 'no more than four questions' is just the opposite 'fail the first four questions' and the probability of one of them is 1 minus the probability of the other. So, what is 1-P where P is the probability that the student will fail four times in a row?

Yeah, but why would we want to find out what is the probability
of student failing all 4 times if we want to find what is the probability of
answering at any of the 4 four times the question is asked.
If we compare it to a biased coin. prob of heads=0.4 and
prob of tails=0.6 and try to find prob of getting heads at least once
in 4 trials (standart binomial distribution) we would also count in probs
where heads comes up more than once in 4 trials. But the student is
not asked any more if he gives correct answer the first time. The trials ends as
soon as student gives correct answer, meaning it might be as soon as first question.
What do you think?

xeonas said:

Hello. My problem:
Student can answer each question given to him with a probability of 0.4.
A student is asked till he gives a correct answer.
What is the probability that no more than 4 questions will be given.
If we use P(X)=nC​r​*p^x​​*q^​n−x​​ then p=4? q=0? n=? x=? no idea.
On the other hand (0.4)+(0.6*0.4)+(0.6*0.6*0.4)+(0.6*0.6*0.6*0.4)=0.87,
which is approx 87% might look correct but that's also probably wrong since
such calculation can be done only for independent probabilities and ours is
clearly not independent since later questions depend on answer of previous questions.
Anyway I would greatly appreciate any help.

Click to expand...

On the other hand (0.4)+(0.6*0.4)+(0.6*0.6*0.4)+(0.6*0.6*0.6*0.4)=0. 87 Oh, you solved the problem. Why do you think it is wrong? Why do you think that the answer to the questions are not independent. If it was dependent can the prob of getting it correct always be 0.4?

xeonas said:

Yeah, but why would we want to find out what is the probability
of student failing all 4 times if we want to find what is the probability of
answering at any of the 4 four times the question is asked.
If we compare it to a biased coin. prob of heads=0.4 and
prob of tails=0.6 and try to find prob of getting heads at least once
in 4 trials (standart binomial distribution) we would also count in probs
where heads comes up more than once in 4 trials. But the student is
not asked any more if he gives correct answer the first time. The trials ends as
soon as student gives correct answer, meaning it might be as soon as first question.
What do you think?

Click to expand...

When you say something like "the student is not asked any more if he gives correct answer the first time" that is true as far as it goes but you are trying to calculate a probability which requires a 'number of occurrences of certain event'. So, how many ways can the student answer the question correctly the first time, how about fail the first and answer the question correctly the second time, or fail the first and second time and answer the question correctly the third time, or fail the first, second and third time and answer the question correctly the fourth time and what is the probability of each of those so you can add them up. If you want to do it that way, how about the first 20 times, first 50 times, ...

I would rather compute the probability of failure 4 times (n times) in a row and subtract it from one. First of all, it is much fewer computations and, secondly, it is simpler. 1-0.6⁴ ~ 0.8704 or 87.04%

EDIT: BTW
(0.4)+(0.6*0.4)+(0.6*0.6*0.4)+(0.6*0.6*0.6*0.4)
=\(\displaystyle 0.4 * \Sigma_{n=0}^{n=3} 0.6^n = 0.4 * \frac{1-0.6^4}{1-0.6} = 0.4 * \frac{1-0.6^4}{0.4} = 1 - 0.6^4\)

Yes, I see now. You are both right. Thanks for the answers.

Also maybe someone knows if its binomial distribution or something else.
Would like to find some more examples like this to practice.