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.
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?
((10*9*8*7)/4!) ;/4^10 - Wolfram|Alpha
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'.
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
If you can either prove your formula or direct me to a proof, I would gladly look into using that method instead.
, 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 .
Thus on a test of ten questions with four options, a student just guesses.
The probability of exactly 4 successes in 10 is .
For at least four is
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
Essentially we are saying: 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..?
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
and compare it to your answer in hopes of gaining some insight into your answer, but we get different answers!
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.