Last nights networking course, we got a bonus question?
Question was: what is the probability to solve all the questions answer right in “4 to 4 matching sentence question*“.
My answer was : 1/4 * 1/4 * 1/4 * 1/4 = 1/156 = 0.006410256
My explanation: getting one right is 1/4, getting 2 right in a row will be 1/4 *1/4 and so on 4 them right will be 1/156
Kafi vai’s answer was: 1/24 = 0.041667
Kafi vai’s explanation: there are 24 combination to have the answers. getting all right it one of 24. its 1/24.
on the other hand, if you calculate probability of each part , it slove the problem also:
1/4 X 1/3 X 1/2 X 1/1 = 1/24 = 0.4166
[updated ]I was Doing big mistake using Binomial Distribution, it will not be used here.
this will be used in sequence like, you have 3dice, how many time you will get 9, 23 on single roll of 3dice.
*[there where a table with 2collums and 4rows, you have to match left with right.]
14/05/2010 at 12:30 pm Permalink
4^4=256
his solution is wrong. there are 4^4 possibilities, not 24 because we can select any one of a,b,c,d for each question. That means there may have repetition.
check this site
http://www.mathsisfun.com/combinatorics/combinations-permutations.html
Ur last solution find probability for 1 question to be right, not all 4 questions (from the theory u wrought, i cant realize math).
14/05/2010 at 4:04 pm Permalink
Nafis,
Interesting problem. If I understood the question correctly, then I think 1/24 is correct. Since you have only 24 choices, if you match all the 4 questions to different answers, correctly or incorrectly. (note that it’s not choice of answers, but the choice of the sequence of answers) and only one sequence is correct. So probability of choosing the correct sequence is 1/24.
(e.g. a single sequence may be q1->a1, q2->a3, q3->a4, q4->a2.
[q for questions and a for matching answers]
This should be counted as only one choice, not four; and there are total 24 such choices.)
Also, as far as I know, the Binomial distribution works for probability of random variables (not random sequences). For example, if you roll a dice 4 times, what is the probability of any predetermined sequence to occur. Say the predetermined sequence is 1,4,6,1. So here, each time you roll the dice, you have equal (in other wards, random) probability distribution. But, in case of matching the questions, after choosing the first question, you can choose the second one only from the remaining three and so on.
So this is not probability of a random variable, but probability of a random sequence, which is different. Look into the probability book for “Probability of a Random Sequence”. That should give you the solution.
I’ll let you know if anything else comes up to my mind.
14/05/2010 at 5:21 pm Permalink
@miraz vai:
1. 1/24 will be correct, if it not related with each other. but for being all them correct, each of them is dependent on other.
2. it matching 4 items with another 4, if you have done 3correct then last one will be automatically correct.
So, for the first question you have 1/4 chance for that one to be correct. then for the 2nd question its 1/3, and 1/2 and 1 for the last one. you need binomial distribution. because the probability is changing.
14/05/2010 at 7:24 pm Permalink
Both me and one of my colleague agrees that the answer is 1/24. Explanation: Suppose, the questions are Q1, Q2, Q3 and Q4 and correct answers are A1, A2, A3 and A4 respectively. Fix the sequence Q1Q2Q3Q4. If answers are randomly chosen, the problem simplifies to arranging A1A2A3A4 in all possible ways. There are 4! = 24 possible ways. We seek the probability of matching answers for all questions and there is only 1 sequence: A1A2A3A4. So the answer is 1/24.
We also cannot believe the answer 0.968 to be true like you. If chance of getting ALL correct answers is so close to 1, students would not study for matching left & right type of questions.
Ask your instructor how s/he got so high probability and let us know.
14/05/2010 at 7:39 pm Permalink
Sorry didn’t notice other comments. My ISP is caching pages!
@Nafis
I think there is a mistake is your reasoning. “for the first question you have 1/4 chance for that one to be correct. then for the 2nd question its 1/3″.
For the first question, 1/4 chance that it will be correct.Say, the answer chosen for the first question matches. Given this, there is a 1/3 chance that the answer chosen for 2nd question will be correct.
Say, the answer chosen for the first question doesn’t match. And the chosen answer is actually the correct answer for question 2. Given this, there is a 0 chance that the answer chosen for 2nd question will be correct.
From the link of binomial distribution: “the binomial distribution is the discrete probability distribution of the number of successes in a sequence of n independent yes/no experiments“. The experiments have to be independent. This is the key to apply binomial distribution. But according to the question, experiments are not independent: “the probability is changing.”
15/05/2010 at 10:11 am Permalink
@miraz vai: thankx for the long talk over phone, it solved the problem.
@Nishad: You are right!
1/24 will be the right answer, the probability means: 1 / how many choice you have .. Here I have 24 options.
17/05/2010 at 2:31 pm Permalink
Answer from our teacher:
Nafis,
You got the answer right, which is 1/4*1/4*1/4*1/4.
There are two important assumption made, first one is the answerer has no clue about the right answe,r hence (s)he is unbiased toward any specific answer. Second one is, having one answer right has no bearing on the probability of having another right (or wrong), i.e. each event of giving correct answer is independent.
Lets define an event to be answering one question right.
Now, how many events were there? Obviously 4.
What is the probability of any one of the event occurring? Since there are 4 unbiased options to choose from: 1/4 .
What is the probability of having all 4 events happening? (1/4)*(1/4)*(1/4)*(1/4).
There is another way of getting to the same answer. Think about in how many ways these 4 questions could be answered: for the first question there are 4 possible ways, for the second question again 4 possible ways (this is an important point to be noted and was explicitly mentioned in the class while the exam was going on, that is, any number of question can be plotted onto one answer.), the same is true for 3rd and 4th question. Thus the total number of ways the 4 questions could be answered is 4*4*4*4. Since there is only one combination that is correct, the probability of giving all 4 correct answers (only by dint of luck) is 1/256.
Hope this puts the matter to rest. But keep the blog open, It is fun to watch people argue. It was a joy talking to a bunch of smart kids every week for 4 months. Take care.
Sabbir
12/01/2012 at 4:20 am Permalink
Suddenly stumbled upon this page from Google’s new Search Plus option
Ya, I remember talking to you over phone on this.
Finally: Yes, it’ll be 1/256 if you mention that one answer may to chosen for more than one questions.
However, as normally happens in the exam, if one answer to be matched against only one question (one-to-one relation), then it’ll be 1/24
BTW, your post still says: 1/4 * 1/4 * 1/4 * 1/4 = 1/156
It’l be: 1/4 * 1/4 * 1/4 * 1/4 = 1/256
Correct it