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I dont reallt understand your question !
60 -50 -40 ???
What was the book?
Ah, right !!!
I was looking for expressing the second bit in terms of A and B, as you said, but I wrote it as:
P(Answer correct) = P(A correct and B correct | (A correct and B correct) + (A wrong and B wrong) = 1)
) thus I counld not do anything with it !
Thanks so much mathsyperson You'rs kind !
popo
Hi, Thank you mathsyperson for giving this help. I presented your reasoning to my professor and he said "It seems to me that is correct" . So, many thanks.
But in the same time, he asked me to formalize this reasoning using "conditional probability". I am trying to do this but it's not trivial.
I think, it should be something like this :
P(the answer is correct) = P ( (A = right and B = right ) | knowing that they have done the same answer ).
So, how to formalize this condition?
Thanks
Hi, someone please help me with this problem?
A professor ask a series of question to 2 students, A and B. If the students dont know the answer then they dont answer. If they think they have an answer then they answer.
For any question, A doesnt know if B can answer nor the answer of B (if B can answer). The same for B.
Before the test begins, the probability that an answer of A is good is unknown. The same thing for B.
Now, the test begins. Here is the score-sheet of the 2 students after many many questions:
For A: he has answered to 50 questions, with 40 good answers and 10 wrong. (80% good)
For B: he has answered to 75 questions, with 30 good answers and 45 wrong. ( 40%good)
By some calculus, the professor find that :
1/ The probability that the next answer of A is good is 0.8
2/ The probability that the next answer of B is good is 0.4
Suppose that (1) and (2) are correct.
Now, the professor ask the 2 students one more question. This time, both A and B think they can respond, and by CHANCE they give the same answer.
The question is: what is the probability that this answer, given by both A and B, is a good answer?
Uhmmmm....
I think this probability must line between 0.4 and 0.8 ; maybe (0.4+0.8)/2 = 0.6 ???
Thank you in advance
Popo
Hi everybody, it's Popo I've just joined the forum, I'm in the IT field but I'm beginning to learn about Probability and Statistics . Hope to get some help from you .......
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