Stuck at an assignment for conditional probability [duplicate]
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This question already has an answer here:
Probability of student passing an exam
1 answer
So I'm having this assignment in probability regarding conditional probability that states the following:
Each of 25 exams papers has 2 questions written on it. Neither of the 50 questions repeats itself. The student knows the answer for 44 questions. In order for the student to pass the exam he must answer correctly either on the two questions for the paper he has chosen or on one question on the first paper he has chosen and on the first question on the second paper he has chosen.
What is the probability that the student will pass the exam?
Any ideas?
Any sort of help will be appreciated.
Thank you in advance!
probability
asked Jan 8 at 22:59
David MathersDavid Mathers
153
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marked as duplicate by Lord Shark the Unknown, KReiser, Lee David Chung Lin, José Carlos Santos, Cesareo Jan 9 at 10:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Probability of student passing an exam
1 answer
So I'm having this assignment in probability regarding conditional probability that states the following:
Each of 25 exams papers has 2 questions written on it. Neither of the 50 questions repeats itself. The student knows the answer for 44 questions. In order for the student to pass the exam he must answer correctly either on the two questions for the paper he has chosen or on one question on the first paper he has chosen and on the first question on the second paper he has chosen.
What is the probability that the student will pass the exam?
Any ideas?
Any sort of help will be appreciated.
Thank you in advance!
probability
asked Jan 8 at 22:59
David MathersDavid Mathers
153
$endgroup$
marked as duplicate by Lord Shark the Unknown, KReiser, Lee David Chung Lin, José Carlos Santos, Cesareo Jan 9 at 10:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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can you clarify the part about choosing 2 papers ?
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– T. Fo
Jan 8 at 23:01
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Also one part says "the paper he has chosen" (suggesting only one paper is chosen) while the very next line seems to contradict this by having the student choose 2 papers.
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– Michael
Jan 8 at 23:05
add a comment |
$begingroup$
This question already has an answer here:
Probability of student passing an exam
1 answer
So I'm having this assignment in probability regarding conditional probability that states the following:
Each of 25 exams papers has 2 questions written on it. Neither of the 50 questions repeats itself. The student knows the answer for 44 questions. In order for the student to pass the exam he must answer correctly either on the two questions for the paper he has chosen or on one question on the first paper he has chosen and on the first question on the second paper he has chosen.
What is the probability that the student will pass the exam?
Any ideas?
Any sort of help will be appreciated.
Thank you in advance!
probability
asked Jan 8 at 22:59
David MathersDavid Mathers
153
$endgroup$
This question already has an answer here:
Probability of student passing an exam
1 answer
So I'm having this assignment in probability regarding conditional probability that states the following:
Each of 25 exams papers has 2 questions written on it. Neither of the 50 questions repeats itself. The student knows the answer for 44 questions. In order for the student to pass the exam he must answer correctly either on the two questions for the paper he has chosen or on one question on the first paper he has chosen and on the first question on the second paper he has chosen.
What is the probability that the student will pass the exam?
Any ideas?
Any sort of help will be appreciated.
Thank you in advance!
This question already has an answer here:
Probability of student passing an exam
1 answer
probability
probability
asked Jan 8 at 22:59
David MathersDavid Mathers
153
asked Jan 8 at 22:59
David MathersDavid Mathers
153
asked Jan 8 at 22:59
David MathersDavid Mathers
153
asked Jan 8 at 22:59
David MathersDavid Mathers
153
asked Jan 8 at 22:59
David MathersDavid Mathers
153
153
marked as duplicate by Lord Shark the Unknown, KReiser, Lee David Chung Lin, José Carlos Santos, Cesareo Jan 9 at 10:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Lord Shark the Unknown, KReiser, Lee David Chung Lin, José Carlos Santos, Cesareo Jan 9 at 10:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
can you clarify the part about choosing 2 papers ?
$endgroup$
– T. Fo
Jan 8 at 23:01
$begingroup$
Also one part says "the paper he has chosen" (suggesting only one paper is chosen) while the very next line seems to contradict this by having the student choose 2 papers.
$endgroup$
– Michael
Jan 8 at 23:05
add a comment |
1
$begingroup$
can you clarify the part about choosing 2 papers ?
$endgroup$
– T. Fo
Jan 8 at 23:01
$begingroup$
Also one part says "the paper he has chosen" (suggesting only one paper is chosen) while the very next line seems to contradict this by having the student choose 2 papers.
$endgroup$
– Michael
Jan 8 at 23:05
1
1
$begingroup$
can you clarify the part about choosing 2 papers ?
$endgroup$
– T. Fo
Jan 8 at 23:01
$begingroup$
can you clarify the part about choosing 2 papers ?
$endgroup$
– T. Fo
Jan 8 at 23:01
$begingroup$
Also one part says "the paper he has chosen" (suggesting only one paper is chosen) while the very next line seems to contradict this by having the student choose 2 papers.
$endgroup$
– Michael
Jan 8 at 23:05
$begingroup$
Also one part says "the paper he has chosen" (suggesting only one paper is chosen) while the very next line seems to contradict this by having the student choose 2 papers.
$endgroup$
– Michael
Jan 8 at 23:05
add a comment |
1 Answer
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votes
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Here is my understanding of scheme.
If student can't answer both 1st and 2nd questions on paper (case 0), he fails.
If student can't answer 1 question but can answer the other (case 1), he will be given another paper.
If student can't answer at least 1 question on additional paper(case 2), he fails.
Win otherwise.
$$
P(text{case 0}) = frac{C_{44}^0 C_6^2}{C_{50}^2}\
P(text{case 1}) = frac{C_{44}^1 C_6^1}{C_{50}^2}\
P(text{case 2}) = frac{C_{43}^0 C_5^2}{C_{48}^2}\
$$
$P(win) = 1 - (P_0 + P_1 P_2) approx 0.985845$ (that much, maybe I am wrong with exam scheme? UPD1: But if student can take 2 papers regardless his ability to answer to the both questions in 1st paper, chances are even higher, every student should take 2 papers and select 2 questions of 4)
*$C_n^k equiv$ Binomial[n,k]
UPD2:
After posting this answer I mentioned (look at right side menu more often!) that this is duplicate ☹
answered Jan 9 at 4:09
Mike_Mike_
616
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is my understanding of scheme.
If student can't answer both 1st and 2nd questions on paper (case 0), he fails.
If student can't answer 1 question but can answer the other (case 1), he will be given another paper.
If student can't answer at least 1 question on additional paper(case 2), he fails.
Win otherwise.
$$
P(text{case 0}) = frac{C_{44}^0 C_6^2}{C_{50}^2}\
P(text{case 1}) = frac{C_{44}^1 C_6^1}{C_{50}^2}\
P(text{case 2}) = frac{C_{43}^0 C_5^2}{C_{48}^2}\
$$
$P(win) = 1 - (P_0 + P_1 P_2) approx 0.985845$ (that much, maybe I am wrong with exam scheme? UPD1: But if student can take 2 papers regardless his ability to answer to the both questions in 1st paper, chances are even higher, every student should take 2 papers and select 2 questions of 4)
*$C_n^k equiv$ Binomial[n,k]
UPD2:
After posting this answer I mentioned (look at right side menu more often!) that this is duplicate ☹
answered Jan 9 at 4:09
Mike_Mike_
616
$endgroup$
add a comment |
$begingroup$
Here is my understanding of scheme.
If student can't answer both 1st and 2nd questions on paper (case 0), he fails.
If student can't answer 1 question but can answer the other (case 1), he will be given another paper.
If student can't answer at least 1 question on additional paper(case 2), he fails.
Win otherwise.
$$
P(text{case 0}) = frac{C_{44}^0 C_6^2}{C_{50}^2}\
P(text{case 1}) = frac{C_{44}^1 C_6^1}{C_{50}^2}\
P(text{case 2}) = frac{C_{43}^0 C_5^2}{C_{48}^2}\
$$
$P(win) = 1 - (P_0 + P_1 P_2) approx 0.985845$ (that much, maybe I am wrong with exam scheme? UPD1: But if student can take 2 papers regardless his ability to answer to the both questions in 1st paper, chances are even higher, every student should take 2 papers and select 2 questions of 4)
*$C_n^k equiv$ Binomial[n,k]
UPD2:
After posting this answer I mentioned (look at right side menu more often!) that this is duplicate ☹
answered Jan 9 at 4:09
Mike_Mike_
616
$endgroup$
add a comment |
$begingroup$
Here is my understanding of scheme.
If student can't answer both 1st and 2nd questions on paper (case 0), he fails.
If student can't answer 1 question but can answer the other (case 1), he will be given another paper.
If student can't answer at least 1 question on additional paper(case 2), he fails.
Win otherwise.
$$
P(text{case 0}) = frac{C_{44}^0 C_6^2}{C_{50}^2}\
P(text{case 1}) = frac{C_{44}^1 C_6^1}{C_{50}^2}\
P(text{case 2}) = frac{C_{43}^0 C_5^2}{C_{48}^2}\
$$
$P(win) = 1 - (P_0 + P_1 P_2) approx 0.985845$ (that much, maybe I am wrong with exam scheme? UPD1: But if student can take 2 papers regardless his ability to answer to the both questions in 1st paper, chances are even higher, every student should take 2 papers and select 2 questions of 4)
*$C_n^k equiv$ Binomial[n,k]
UPD2:
After posting this answer I mentioned (look at right side menu more often!) that this is duplicate ☹
answered Jan 9 at 4:09
Mike_Mike_
616
$endgroup$
Here is my understanding of scheme.
If student can't answer both 1st and 2nd questions on paper (case 0), he fails.
If student can't answer 1 question but can answer the other (case 1), he will be given another paper.
If student can't answer at least 1 question on additional paper(case 2), he fails.
Win otherwise.
$$
P(text{case 0}) = frac{C_{44}^0 C_6^2}{C_{50}^2}\
P(text{case 1}) = frac{C_{44}^1 C_6^1}{C_{50}^2}\
P(text{case 2}) = frac{C_{43}^0 C_5^2}{C_{48}^2}\
$$
$P(win) = 1 - (P_0 + P_1 P_2) approx 0.985845$ (that much, maybe I am wrong with exam scheme? UPD1: But if student can take 2 papers regardless his ability to answer to the both questions in 1st paper, chances are even higher, every student should take 2 papers and select 2 questions of 4)
*$C_n^k equiv$ Binomial[n,k]
UPD2:
After posting this answer I mentioned (look at right side menu more often!) that this is duplicate ☹
answered Jan 9 at 4:09
Mike_Mike_
616
edited Jan 9 at 4:39
answered Jan 9 at 4:09
Mike_Mike_
616
answered Jan 9 at 4:09
Mike_Mike_
616
answered Jan 9 at 4:09
Mike_Mike_
616
616
add a comment |
add a comment |
1
$begingroup$
can you clarify the part about choosing 2 papers ?
$endgroup$
– T. Fo
Jan 8 at 23:01
$begingroup$
Also one part says "the paper he has chosen" (suggesting only one paper is chosen) while the very next line seems to contradict this by having the student choose 2 papers.
$endgroup$
– Michael
Jan 8 at 23:05