1. ## Probability problems

A test contains 10 questions and each question has 4 options to answer in with 1 being the correct answer.
What is the probability of getting at least 4 questions right if answers are randomly selected?

What is the probability of getting between 2 to 7 questions right (3,4,5,6) if answers are randomly selected?

I'm completely stranded in probability. I'm trying to get it in the P and C formats but it won't work. Help is very much appreciated.

2. ## Re: Probability problems

Originally Posted by Paze
[I]A test contains 10 questions and each question has 4 options to answer in with 1 being the correct answer.
What is the probability of getting at least 4 questions right if answers are randomly selected?

What is the probability of getting between 2 to 7 questions right (3,4,5,6) if answers are randomly selected?

The probability of getting exactly k correct is:
$\displaystyle \mathcal{P}(X=k)=\binom{10}{k}(.25)^k(.75)^{10-k}$

Now just express these as sums.

3. ## Re: Probability problems

Thank you. Is this the correct answer to my first problem?

((10*9*8*7*6*5*4)/7!)/4^10 - Wolfram|Alpha

And this for my 2nd?

&#40;&#40;10&#42;9&#42;8&#42;7&#41;&#47;4&#33;&#41 ;&#47;4&#94;10 - Wolfram|Alpha

4. ## Re: Probability problems

Originally Posted by Paze
Thank you. Is this the correct answer to my first problem?
((10*9*8*7*6*5*4)/7!)/4^10 - Wolfram|Alpha
NO!
The answer to the first is $\displaystyle \sum\limits_{k = 4}^{10} {\mathcal{P}(X=k)}$

5. ## Re: Probability problems

Originally Posted by Plato
NO!
The answer to the first is $\displaystyle \sum\limits_{k = 4}^{10} {\mathcal{P}(X=k)}$
Hm, I didn't learn it with sums

Is that the only way to do it?

I feel like this is the same problem: https://www.khanacademy.org/math/pro...-in-five-flips

Only my problem contains 4 possibilities instead of 2 and I have more 'flips'.

6. ## Re: Probability problems

For the first, you break it down into cases and use the addition principle.

Case 1: Exactly 4 questions right
-Pick 4 questions to be right. The other 6 are wrong. What is the probability of this?

Case 2: Exactly 5 questions right
and so on.

The 2nd problem uses cases and addition principle too, however the cases are different.

This isn't straight up counting per se , but independent trails that are either a success with probability .25 or a failure with probability .75

7. ## Re: Probability problems

Originally Posted by Paze
I feel like this is the same problem: https://www.khanacademy.org/math/pro...-in-five-flips
Did you read reply #2? That shows you exactly how to do these.

Now if you are foolish enough to use KhanAcademy for serious study, then you get what you deserve.

8. ## Re: Probability problems

Originally Posted by Plato
Did you read reply #2? That shows you exactly how to do these.

Now if you are foolish enough to use KhanAcademy for serious study, then you get what you deserve.
I did read your reply but I don't even understand it. I'm learning to solve these problems using factorials. I think your way of solving it is more like the binomial theorem while I am asking for pascals triangle..I want to learn the basics before I simply apply a formula to solve it.

If you can either prove your formula or direct me to a proof, I would gladly look into using that method instead.

9. ## Re: Probability problems

Originally Posted by Paze
I did read your reply but I don't even understand it. I'm learning to solve these problems using factorials. I think your way of solving it is more like the binomial theorem while I am asking for pascals triangle..I want to learn the basics before I simply apply a formula to solve it.
If you can either prove your formula or direct me to a proof, I would gladly look into using that method instead.
Do you know about binomial coefficients?
$\displaystyle \binom{N}{k}=\frac{N!}{k!(N-k)!}$, the number of combinations of N objects taken k at a time?

It is also the number of ways the arrange N 1's and k 0's in a string.

We think of the 1's as a place where a success occurs and the 0's where a failure happens.
If success has a probability of p then failure has probability of 1-p.

So the probability of exactly k successes in N trials is $\displaystyle \binom{N}{k}(p)^k(1-p)^{N-k}$.

Thus on a test of ten questions with four options, a student just guesses.
The probability of exactly 4 successes in 10 is $\displaystyle \binom{10}{4}(.25)^4(.75)^{6}$.

For at least four is
$\displaystyle \binom{10}{4}(.25)^4(.75)^{6}+\binom{10}{5}(.25)^4 (.75)^{6}+\binom{10}{7}(.25)^7(.75)^{3}+\binom{10} {8}(.25)^8(.75)^{2}+\binom{10}{9}(.25)^9(.75)^{1}+ \binom{10}{10}(.25)^{10}(.75)^{0}$.

10. ## Re: Probability problems

Originally Posted by Plato
Do you know about binomial coefficients?
$\displaystyle \binom{N}{k}=\frac{N!}{k!(N-k)!}$, the number of combinations of N objects taken k at a time?

It is also the number of ways the arrange N 1's and k 0's in a string.

We think of the 1's as a place where a success occurs and the 0's where a failure happens.
If success has a probability of p then failure has probability of 1-p.

So the probability of exactly k successes in N trials is $\displaystyle \binom{N}{k}(p)^k(1-p)^{N-k}$.

Thus on a test of ten questions with four options, a student just guesses.
The probability of exactly 4 successes in 10 is $\displaystyle \binom{10}{4}(.25)^4(.75)^{6}$.

For at least four is
$\displaystyle \binom{10}{4}(.25)^4(.75)^{6}+\binom{10}{5}(.25)^4 (.75)^{6}+\binom{10}{7}(.25)^7(.75)^{3}+\binom{10} {8}(.25)^8(.75)^{2}+\binom{10}{9}(.25)^9(.75)^{1}+ \binom{10}{10}(.25)^{10}(.75)^{0}$.
Thanks a lot for that. This is actually the kind of logic I have been looking for in probability.

I have come as far as understanding how the 1's represent success and the 0's represent failure. I also understand that if the probability of success is P then the probability of failure is 1-P.

I am trying to understand the formula $\displaystyle \binom{N}{k}(p)^k(1-p)^{N-k}$

Essentially we are saying: $\displaystyle \binom{N}{k}=\frac{N!}{k!(N-k)!}$ which represents what exactly? Is 10 my N and 4 my K? 4 answers (choices) with 10 events (iterations)? In that case does this binomial coefficient represent the total amount of successes and failures and the rest of the formula acts as some sort of filter to answer the rest of the question, e.g. probability of exactly 4 instead of 10..?

11. ## Re: Probability problems

Why isn't this essentially the same as saying:

I have a dice with 4 sides. I throw it 10 times. What is the probability of getting the same side exactly twice?

I calculate it according to https://www.khanacademy.org/math/pro...-combinatorics

12. ## Re: Probability problems

In the dice example each side has an equally likely chance of occuring.

With the test problem, we have two things in the sample space
Right with probability .25
Wrong with probability .75

the outcomes are not equally likely

EDIT: Also is the dice/coin problem, order is important.

In the test problem, order DOES NOT matter.

Getting questions 1,4,6,10 wrong is the same as getting questions 4,10,6,1 wrong. We're not ordering the questions. We just want to pick 4 to be right, and the other 6 wil be wrong.

13. ## Re: Probability problems

Okay, I think this question might have been a bit out of my reach. I will go back to some more basic probability questions and try this one later!