Mixing
Probability of at least 6 out of 10
$begingroup$
i have the following question:
In an exam, 10 questions are chosen randomly from 100.
A student is prepared for 60 out 100 of these questions.
What's the probability that he'll solve at least 6 questions out of the chosen 10?
I'm a bit confused as to what the probability plain and the events are.
Do i say that $|Omega| = {100choose10} $, and the event $A=$"at least 6 of the 10 questions are among the students' 60 " ?
probability combinatorics probability-theory
asked Mar 31 '17 at 9:48
CodeHoarderCodeHoarder
18610
$endgroup$
add a comment |
$begingroup$
i have the following question:
In an exam, 10 questions are chosen randomly from 100.
A student is prepared for 60 out 100 of these questions.
What's the probability that he'll solve at least 6 questions out of the chosen 10?
I'm a bit confused as to what the probability plain and the events are.
Do i say that $|Omega| = {100choose10} $, and the event $A=$"at least 6 of the 10 questions are among the students' 60 " ?
probability combinatorics probability-theory
asked Mar 31 '17 at 9:48
CodeHoarderCodeHoarder
18610
$endgroup$
1
$begingroup$
Yes, see this link on the hypergeometric distribution.
$endgroup$
– N. Shales
Mar 31 '17 at 10:42
$begingroup$
I don't see the connection, sorry.
$endgroup$
– CodeHoarder
Mar 31 '17 at 14:56
1
$begingroup$
Separate the $100$ questions into the $60$ the student's prepared for and the $40$ he hasn't, then there are $binom{100}{10}$ possible question sets. Out of these there are $binom{60}{6}binom{40}{4}$ ways that $6$ will be from the $60$ and $4$ will be from the $40$, thus the probability that the student will solve exactly $6$ questions is $$frac{binom{60}{6}binom{40}{4}}{binom{100}{10}}$$ assuming he solves every question he has prepared for. Similarly the probability that he has prepared for exactly $7$ questions is $$frac{binom{60}{7}binom{40}{3}}{binom{100}{10}} $$ and so on.
$endgroup$
– N. Shales
Mar 31 '17 at 15:20
add a comment |
$begingroup$
i have the following question:
In an exam, 10 questions are chosen randomly from 100.
A student is prepared for 60 out 100 of these questions.
What's the probability that he'll solve at least 6 questions out of the chosen 10?
I'm a bit confused as to what the probability plain and the events are.
Do i say that $|Omega| = {100choose10} $, and the event $A=$"at least 6 of the 10 questions are among the students' 60 " ?
probability combinatorics probability-theory
asked Mar 31 '17 at 9:48
CodeHoarderCodeHoarder
18610
$endgroup$
i have the following question:
In an exam, 10 questions are chosen randomly from 100.
A student is prepared for 60 out 100 of these questions.
What's the probability that he'll solve at least 6 questions out of the chosen 10?
I'm a bit confused as to what the probability plain and the events are.
Do i say that $|Omega| = {100choose10} $, and the event $A=$"at least 6 of the 10 questions are among the students' 60 " ?
probability combinatorics probability-theory
probability combinatorics probability-theory
asked Mar 31 '17 at 9:48
CodeHoarderCodeHoarder
18610
asked Mar 31 '17 at 9:48
CodeHoarderCodeHoarder
18610
asked Mar 31 '17 at 9:48
CodeHoarderCodeHoarder
18610
asked Mar 31 '17 at 9:48
CodeHoarderCodeHoarder
18610
asked Mar 31 '17 at 9:48
CodeHoarderCodeHoarder
18610
18610
1
$begingroup$
Yes, see this link on the hypergeometric distribution.
$endgroup$
– N. Shales
Mar 31 '17 at 10:42
$begingroup$
I don't see the connection, sorry.
$endgroup$
– CodeHoarder
Mar 31 '17 at 14:56
1
$begingroup$
Separate the $100$ questions into the $60$ the student's prepared for and the $40$ he hasn't, then there are $binom{100}{10}$ possible question sets. Out of these there are $binom{60}{6}binom{40}{4}$ ways that $6$ will be from the $60$ and $4$ will be from the $40$, thus the probability that the student will solve exactly $6$ questions is $$frac{binom{60}{6}binom{40}{4}}{binom{100}{10}}$$ assuming he solves every question he has prepared for. Similarly the probability that he has prepared for exactly $7$ questions is $$frac{binom{60}{7}binom{40}{3}}{binom{100}{10}} $$ and so on.
$endgroup$
– N. Shales
Mar 31 '17 at 15:20
add a comment |
1
$begingroup$
Yes, see this link on the hypergeometric distribution.
$endgroup$
– N. Shales
Mar 31 '17 at 10:42
$begingroup$
I don't see the connection, sorry.
$endgroup$
– CodeHoarder
Mar 31 '17 at 14:56
1
$begingroup$
Separate the $100$ questions into the $60$ the student's prepared for and the $40$ he hasn't, then there are $binom{100}{10}$ possible question sets. Out of these there are $binom{60}{6}binom{40}{4}$ ways that $6$ will be from the $60$ and $4$ will be from the $40$, thus the probability that the student will solve exactly $6$ questions is $$frac{binom{60}{6}binom{40}{4}}{binom{100}{10}}$$ assuming he solves every question he has prepared for. Similarly the probability that he has prepared for exactly $7$ questions is $$frac{binom{60}{7}binom{40}{3}}{binom{100}{10}} $$ and so on.
$endgroup$
– N. Shales
Mar 31 '17 at 15:20
1
1
$begingroup$
Yes, see this link on the hypergeometric distribution.
$endgroup$
– N. Shales
Mar 31 '17 at 10:42
$begingroup$
Yes, see this link on the hypergeometric distribution.
$endgroup$
– N. Shales
Mar 31 '17 at 10:42
$begingroup$
I don't see the connection, sorry.
$endgroup$
– CodeHoarder
Mar 31 '17 at 14:56
$begingroup$
I don't see the connection, sorry.
$endgroup$
– CodeHoarder
Mar 31 '17 at 14:56
1
1
$begingroup$
Separate the $100$ questions into the $60$ the student's prepared for and the $40$ he hasn't, then there are $binom{100}{10}$ possible question sets. Out of these there are $binom{60}{6}binom{40}{4}$ ways that $6$ will be from the $60$ and $4$ will be from the $40$, thus the probability that the student will solve exactly $6$ questions is $$frac{binom{60}{6}binom{40}{4}}{binom{100}{10}}$$ assuming he solves every question he has prepared for. Similarly the probability that he has prepared for exactly $7$ questions is $$frac{binom{60}{7}binom{40}{3}}{binom{100}{10}} $$ and so on.
$endgroup$
– N. Shales
Mar 31 '17 at 15:20
$begingroup$
Separate the $100$ questions into the $60$ the student's prepared for and the $40$ he hasn't, then there are $binom{100}{10}$ possible question sets. Out of these there are $binom{60}{6}binom{40}{4}$ ways that $6$ will be from the $60$ and $4$ will be from the $40$, thus the probability that the student will solve exactly $6$ questions is $$frac{binom{60}{6}binom{40}{4}}{binom{100}{10}}$$ assuming he solves every question he has prepared for. Similarly the probability that he has prepared for exactly $7$ questions is $$frac{binom{60}{7}binom{40}{3}}{binom{100}{10}} $$ and so on.
$endgroup$
– N. Shales
Mar 31 '17 at 15:20
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A lot of probability problems can be solved by reframing the question in the context of balls and urns. In this case, the hypergeometric distribution is pertinent as the following excerpt from Blitzstein shows:
Consider an urn with $w$ white balls and $b$ black balls. We draw $n$ balls out of the urn at random without replacement, such that all ${w+b choose n}$ samples are equally likely. Let $X$ be the number of white balls in the sample. Then $X$ is said to have the Hypergeometric distribution with parameters $w$, $b$, and $n$; we denote this by $X sim text{HGeom}(w,b,n)$.
For your problem, $n$ is the number of questions are chosen randomly, $w$ is the number of questions that the student is prepared for, $b$ is the number of questions he is not prepared for,
and $X$ is the number of correct answers on the student's exam.
Now that you have properly modeled the problem, all you have to do is calculate
$$P(text{at least 6 out of 10}) = P(X geq 6), $$
with the appropriate pmf.
answered Apr 4 '17 at 19:56
AlainAlain
6151513
$endgroup$
add a comment |
$begingroup$
A permutational approach - Equivalence of permutation and combination formula
Define event A to be there are at least 6 successes out of n = 10
trials. Let's also use symbols: "s" for success, "f" for failure, "t"
for trial" "Pr" for probability, "P" for permutation and "C" for
combination,
"P(n, r)" for selecting r out of n objects in permutation and C(n, r)
for select r out of n objects in combination.
Exactly 6 s out of 10 t, assume first 6 are s, last 4 are f "in that order".
Pr($A_6$, po) = 60/100 * 59/99 * 58/98 * .../55/95 * 40/94 * 39/93 * ..* 37/91
= P (60, 6)* P(40, 4)/ P(100, 10)
But there are C (10, 6) = P(10, 6)/ 6! = I such possibilities.
So exactly 6 s out of 10 t "in any order" is
Pr ($A_6$) = $sum_{ i =1}^I$Pr ($A_6$, po) = I * Pr($A_6$, po)
= C(10, 6) * P(60, 6) * P(40, 4) / P(100, 10)
= C(10, 6) * (6!* 4! /10!) *C(60, 6) * C(40, 4) / C(100,10)
= C(10, 6)/C(10,6) * C(60, 6) * C(40, 4) / C(100,10)
= C(60, 6) * C(40, 4) / C(100,10),
which is exactly the results using combination approach!
The probability of solving at least 6 s out of 10 t is
Pr(A) = $sum_{i=6}^{10}$ Pr($A_i$).
answered Mar 31 '17 at 12:16
cdeamazecdeamaze
114
$endgroup$
$begingroup$
This solution assumes independence of selection. But as (presumably) a given question won't be selected more than once, we wouldn't have independence.
$endgroup$
– paw88789
Mar 31 '17 at 12:24
$begingroup$
Thanks, you have a very good point! Instead of p^ k, do I need p1= P- i 1)/(N-I+))
$endgroup$
– cdeamaze
Mar 31 '17 at 13:36
$begingroup$
Sorry, this is unclear to me.
$endgroup$
– CodeHoarder
Mar 31 '17 at 14:54
$begingroup$
I will have to come back at a later time to post an update. Sorry, my post was incomplete as I hit a return button by mistake. When I tried to correct it, I was told to wait until at least tomorrow as I have used my daily quota which is one post a day. I am a newbie, so some restrictions applied. Stay tuned.
$endgroup$
– cdeamaze
Mar 31 '17 at 22:32
$begingroup$
I am still not allowed to post.
$endgroup$
– cdeamaze
Apr 3 '17 at 3:56
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A lot of probability problems can be solved by reframing the question in the context of balls and urns. In this case, the hypergeometric distribution is pertinent as the following excerpt from Blitzstein shows:
Consider an urn with $w$ white balls and $b$ black balls. We draw $n$ balls out of the urn at random without replacement, such that all ${w+b choose n}$ samples are equally likely. Let $X$ be the number of white balls in the sample. Then $X$ is said to have the Hypergeometric distribution with parameters $w$, $b$, and $n$; we denote this by $X sim text{HGeom}(w,b,n)$.
For your problem, $n$ is the number of questions are chosen randomly, $w$ is the number of questions that the student is prepared for, $b$ is the number of questions he is not prepared for,
and $X$ is the number of correct answers on the student's exam.
Now that you have properly modeled the problem, all you have to do is calculate
$$P(text{at least 6 out of 10}) = P(X geq 6), $$
with the appropriate pmf.
answered Apr 4 '17 at 19:56
AlainAlain
6151513
$endgroup$
add a comment |
$begingroup$
A lot of probability problems can be solved by reframing the question in the context of balls and urns. In this case, the hypergeometric distribution is pertinent as the following excerpt from Blitzstein shows:
Consider an urn with $w$ white balls and $b$ black balls. We draw $n$ balls out of the urn at random without replacement, such that all ${w+b choose n}$ samples are equally likely. Let $X$ be the number of white balls in the sample. Then $X$ is said to have the Hypergeometric distribution with parameters $w$, $b$, and $n$; we denote this by $X sim text{HGeom}(w,b,n)$.
For your problem, $n$ is the number of questions are chosen randomly, $w$ is the number of questions that the student is prepared for, $b$ is the number of questions he is not prepared for,
and $X$ is the number of correct answers on the student's exam.
Now that you have properly modeled the problem, all you have to do is calculate
$$P(text{at least 6 out of 10}) = P(X geq 6), $$
with the appropriate pmf.
answered Apr 4 '17 at 19:56
AlainAlain
6151513
$endgroup$
add a comment |
$begingroup$
A lot of probability problems can be solved by reframing the question in the context of balls and urns. In this case, the hypergeometric distribution is pertinent as the following excerpt from Blitzstein shows:
Consider an urn with $w$ white balls and $b$ black balls. We draw $n$ balls out of the urn at random without replacement, such that all ${w+b choose n}$ samples are equally likely. Let $X$ be the number of white balls in the sample. Then $X$ is said to have the Hypergeometric distribution with parameters $w$, $b$, and $n$; we denote this by $X sim text{HGeom}(w,b,n)$.
For your problem, $n$ is the number of questions are chosen randomly, $w$ is the number of questions that the student is prepared for, $b$ is the number of questions he is not prepared for,
and $X$ is the number of correct answers on the student's exam.
Now that you have properly modeled the problem, all you have to do is calculate
$$P(text{at least 6 out of 10}) = P(X geq 6), $$
with the appropriate pmf.
answered Apr 4 '17 at 19:56
AlainAlain
6151513
$endgroup$
A lot of probability problems can be solved by reframing the question in the context of balls and urns. In this case, the hypergeometric distribution is pertinent as the following excerpt from Blitzstein shows:
Consider an urn with $w$ white balls and $b$ black balls. We draw $n$ balls out of the urn at random without replacement, such that all ${w+b choose n}$ samples are equally likely. Let $X$ be the number of white balls in the sample. Then $X$ is said to have the Hypergeometric distribution with parameters $w$, $b$, and $n$; we denote this by $X sim text{HGeom}(w,b,n)$.
For your problem, $n$ is the number of questions are chosen randomly, $w$ is the number of questions that the student is prepared for, $b$ is the number of questions he is not prepared for,
and $X$ is the number of correct answers on the student's exam.
Now that you have properly modeled the problem, all you have to do is calculate
$$P(text{at least 6 out of 10}) = P(X geq 6), $$
with the appropriate pmf.
answered Apr 4 '17 at 19:56
AlainAlain
6151513
answered Apr 4 '17 at 19:56
AlainAlain
6151513
answered Apr 4 '17 at 19:56
AlainAlain
6151513
answered Apr 4 '17 at 19:56
AlainAlain
6151513
6151513
add a comment |
add a comment |
$begingroup$
A permutational approach - Equivalence of permutation and combination formula
Define event A to be there are at least 6 successes out of n = 10
trials. Let's also use symbols: "s" for success, "f" for failure, "t"
for trial" "Pr" for probability, "P" for permutation and "C" for
combination,
"P(n, r)" for selecting r out of n objects in permutation and C(n, r)
for select r out of n objects in combination.
Exactly 6 s out of 10 t, assume first 6 are s, last 4 are f "in that order".
Pr($A_6$, po) = 60/100 * 59/99 * 58/98 * .../55/95 * 40/94 * 39/93 * ..* 37/91
= P (60, 6)* P(40, 4)/ P(100, 10)
But there are C (10, 6) = P(10, 6)/ 6! = I such possibilities.
So exactly 6 s out of 10 t "in any order" is
Pr ($A_6$) = $sum_{ i =1}^I$Pr ($A_6$, po) = I * Pr($A_6$, po)
= C(10, 6) * P(60, 6) * P(40, 4) / P(100, 10)
= C(10, 6) * (6!* 4! /10!) *C(60, 6) * C(40, 4) / C(100,10)
= C(10, 6)/C(10,6) * C(60, 6) * C(40, 4) / C(100,10)
= C(60, 6) * C(40, 4) / C(100,10),
which is exactly the results using combination approach!
The probability of solving at least 6 s out of 10 t is
Pr(A) = $sum_{i=6}^{10}$ Pr($A_i$).
answered Mar 31 '17 at 12:16
cdeamazecdeamaze
114
$endgroup$
$begingroup$
This solution assumes independence of selection. But as (presumably) a given question won't be selected more than once, we wouldn't have independence.
$endgroup$
– paw88789
Mar 31 '17 at 12:24
$begingroup$
Thanks, you have a very good point! Instead of p^ k, do I need p1= P- i 1)/(N-I+))
$endgroup$
– cdeamaze
Mar 31 '17 at 13:36
$begingroup$
Sorry, this is unclear to me.
$endgroup$
– CodeHoarder
Mar 31 '17 at 14:54
$begingroup$
I will have to come back at a later time to post an update. Sorry, my post was incomplete as I hit a return button by mistake. When I tried to correct it, I was told to wait until at least tomorrow as I have used my daily quota which is one post a day. I am a newbie, so some restrictions applied. Stay tuned.
$endgroup$
– cdeamaze
Mar 31 '17 at 22:32
$begingroup$
I am still not allowed to post.
$endgroup$
– cdeamaze
Apr 3 '17 at 3:56
add a comment |
$begingroup$
A permutational approach - Equivalence of permutation and combination formula
Define event A to be there are at least 6 successes out of n = 10
trials. Let's also use symbols: "s" for success, "f" for failure, "t"
for trial" "Pr" for probability, "P" for permutation and "C" for
combination,
"P(n, r)" for selecting r out of n objects in permutation and C(n, r)
for select r out of n objects in combination.
Exactly 6 s out of 10 t, assume first 6 are s, last 4 are f "in that order".
Pr($A_6$, po) = 60/100 * 59/99 * 58/98 * .../55/95 * 40/94 * 39/93 * ..* 37/91
= P (60, 6)* P(40, 4)/ P(100, 10)
But there are C (10, 6) = P(10, 6)/ 6! = I such possibilities.
So exactly 6 s out of 10 t "in any order" is
Pr ($A_6$) = $sum_{ i =1}^I$Pr ($A_6$, po) = I * Pr($A_6$, po)
= C(10, 6) * P(60, 6) * P(40, 4) / P(100, 10)
= C(10, 6) * (6!* 4! /10!) *C(60, 6) * C(40, 4) / C(100,10)
= C(10, 6)/C(10,6) * C(60, 6) * C(40, 4) / C(100,10)
= C(60, 6) * C(40, 4) / C(100,10),
which is exactly the results using combination approach!
The probability of solving at least 6 s out of 10 t is
Pr(A) = $sum_{i=6}^{10}$ Pr($A_i$).
answered Mar 31 '17 at 12:16
cdeamazecdeamaze
114
$endgroup$
$begingroup$
This solution assumes independence of selection. But as (presumably) a given question won't be selected more than once, we wouldn't have independence.
$endgroup$
– paw88789
Mar 31 '17 at 12:24
$begingroup$
Thanks, you have a very good point! Instead of p^ k, do I need p1= P- i 1)/(N-I+))
$endgroup$
– cdeamaze
Mar 31 '17 at 13:36
$begingroup$
Sorry, this is unclear to me.
$endgroup$
– CodeHoarder
Mar 31 '17 at 14:54
$begingroup$
I will have to come back at a later time to post an update. Sorry, my post was incomplete as I hit a return button by mistake. When I tried to correct it, I was told to wait until at least tomorrow as I have used my daily quota which is one post a day. I am a newbie, so some restrictions applied. Stay tuned.
$endgroup$
– cdeamaze
Mar 31 '17 at 22:32
$begingroup$
I am still not allowed to post.
$endgroup$
– cdeamaze
Apr 3 '17 at 3:56
add a comment |
$begingroup$
A permutational approach - Equivalence of permutation and combination formula
Define event A to be there are at least 6 successes out of n = 10
trials. Let's also use symbols: "s" for success, "f" for failure, "t"
for trial" "Pr" for probability, "P" for permutation and "C" for
combination,
"P(n, r)" for selecting r out of n objects in permutation and C(n, r)
for select r out of n objects in combination.
Exactly 6 s out of 10 t, assume first 6 are s, last 4 are f "in that order".
Pr($A_6$, po) = 60/100 * 59/99 * 58/98 * .../55/95 * 40/94 * 39/93 * ..* 37/91
= P (60, 6)* P(40, 4)/ P(100, 10)
But there are C (10, 6) = P(10, 6)/ 6! = I such possibilities.
So exactly 6 s out of 10 t "in any order" is
Pr ($A_6$) = $sum_{ i =1}^I$Pr ($A_6$, po) = I * Pr($A_6$, po)
= C(10, 6) * P(60, 6) * P(40, 4) / P(100, 10)
= C(10, 6) * (6!* 4! /10!) *C(60, 6) * C(40, 4) / C(100,10)
= C(10, 6)/C(10,6) * C(60, 6) * C(40, 4) / C(100,10)
= C(60, 6) * C(40, 4) / C(100,10),
which is exactly the results using combination approach!
The probability of solving at least 6 s out of 10 t is
Pr(A) = $sum_{i=6}^{10}$ Pr($A_i$).
answered Mar 31 '17 at 12:16
cdeamazecdeamaze
114
$endgroup$
A permutational approach - Equivalence of permutation and combination formula
Define event A to be there are at least 6 successes out of n = 10
trials. Let's also use symbols: "s" for success, "f" for failure, "t"
for trial" "Pr" for probability, "P" for permutation and "C" for
combination,
"P(n, r)" for selecting r out of n objects in permutation and C(n, r)
for select r out of n objects in combination.
Exactly 6 s out of 10 t, assume first 6 are s, last 4 are f "in that order".
Pr($A_6$, po) = 60/100 * 59/99 * 58/98 * .../55/95 * 40/94 * 39/93 * ..* 37/91
= P (60, 6)* P(40, 4)/ P(100, 10)
But there are C (10, 6) = P(10, 6)/ 6! = I such possibilities.
So exactly 6 s out of 10 t "in any order" is
Pr ($A_6$) = $sum_{ i =1}^I$Pr ($A_6$, po) = I * Pr($A_6$, po)
= C(10, 6) * P(60, 6) * P(40, 4) / P(100, 10)
= C(10, 6) * (6!* 4! /10!) *C(60, 6) * C(40, 4) / C(100,10)
= C(10, 6)/C(10,6) * C(60, 6) * C(40, 4) / C(100,10)
= C(60, 6) * C(40, 4) / C(100,10),
which is exactly the results using combination approach!
The probability of solving at least 6 s out of 10 t is
Pr(A) = $sum_{i=6}^{10}$ Pr($A_i$).
answered Mar 31 '17 at 12:16
cdeamazecdeamaze
114
edited Apr 5 '17 at 23:02
answered Mar 31 '17 at 12:16
cdeamazecdeamaze
114
answered Mar 31 '17 at 12:16
cdeamazecdeamaze
114
answered Mar 31 '17 at 12:16
cdeamazecdeamaze
114
114
$begingroup$
This solution assumes independence of selection. But as (presumably) a given question won't be selected more than once, we wouldn't have independence.
$endgroup$
– paw88789
Mar 31 '17 at 12:24
$begingroup$
Thanks, you have a very good point! Instead of p^ k, do I need p1= P- i 1)/(N-I+))
$endgroup$
– cdeamaze
Mar 31 '17 at 13:36
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Sorry, this is unclear to me.
$endgroup$
– CodeHoarder
Mar 31 '17 at 14:54
$begingroup$
I will have to come back at a later time to post an update. Sorry, my post was incomplete as I hit a return button by mistake. When I tried to correct it, I was told to wait until at least tomorrow as I have used my daily quota which is one post a day. I am a newbie, so some restrictions applied. Stay tuned.
$endgroup$
– cdeamaze
Mar 31 '17 at 22:32
$begingroup$
I am still not allowed to post.
$endgroup$
– cdeamaze
Apr 3 '17 at 3:56
add a comment |
$begingroup$
This solution assumes independence of selection. But as (presumably) a given question won't be selected more than once, we wouldn't have independence.
$endgroup$
– paw88789
Mar 31 '17 at 12:24
$begingroup$
Thanks, you have a very good point! Instead of p^ k, do I need p1= P- i 1)/(N-I+))
$endgroup$
– cdeamaze
Mar 31 '17 at 13:36
$begingroup$
Sorry, this is unclear to me.
$endgroup$
– CodeHoarder
Mar 31 '17 at 14:54
$begingroup$
I will have to come back at a later time to post an update. Sorry, my post was incomplete as I hit a return button by mistake. When I tried to correct it, I was told to wait until at least tomorrow as I have used my daily quota which is one post a day. I am a newbie, so some restrictions applied. Stay tuned.
$endgroup$
– cdeamaze
Mar 31 '17 at 22:32
$begingroup$
I am still not allowed to post.
$endgroup$
– cdeamaze
Apr 3 '17 at 3:56
$begingroup$
This solution assumes independence of selection. But as (presumably) a given question won't be selected more than once, we wouldn't have independence.
$endgroup$
– paw88789
Mar 31 '17 at 12:24
$begingroup$
This solution assumes independence of selection. But as (presumably) a given question won't be selected more than once, we wouldn't have independence.
$endgroup$
– paw88789
Mar 31 '17 at 12:24
$begingroup$
Thanks, you have a very good point! Instead of p^ k, do I need p1= P- i 1)/(N-I+))
$endgroup$
– cdeamaze
Mar 31 '17 at 13:36
$begingroup$
Thanks, you have a very good point! Instead of p^ k, do I need p1= P- i 1)/(N-I+))
$endgroup$
– cdeamaze
Mar 31 '17 at 13:36
$begingroup$
Sorry, this is unclear to me.
$endgroup$
– CodeHoarder
Mar 31 '17 at 14:54
$begingroup$
Sorry, this is unclear to me.
$endgroup$
– CodeHoarder
Mar 31 '17 at 14:54
$begingroup$
I will have to come back at a later time to post an update. Sorry, my post was incomplete as I hit a return button by mistake. When I tried to correct it, I was told to wait until at least tomorrow as I have used my daily quota which is one post a day. I am a newbie, so some restrictions applied. Stay tuned.
$endgroup$
– cdeamaze
Mar 31 '17 at 22:32
$begingroup$
I will have to come back at a later time to post an update. Sorry, my post was incomplete as I hit a return button by mistake. When I tried to correct it, I was told to wait until at least tomorrow as I have used my daily quota which is one post a day. I am a newbie, so some restrictions applied. Stay tuned.
$endgroup$
– cdeamaze
Mar 31 '17 at 22:32
$begingroup$
I am still not allowed to post.
$endgroup$
– cdeamaze
Apr 3 '17 at 3:56
$begingroup$
I am still not allowed to post.
$endgroup$
– cdeamaze
Apr 3 '17 at 3:56
add a comment |
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$begingroup$
Yes, see this link on the hypergeometric distribution.
$endgroup$
– N. Shales
Mar 31 '17 at 10:42
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I don't see the connection, sorry.
$endgroup$
– CodeHoarder
Mar 31 '17 at 14:56
1
$begingroup$
Separate the $100$ questions into the $60$ the student's prepared for and the $40$ he hasn't, then there are $binom{100}{10}$ possible question sets. Out of these there are $binom{60}{6}binom{40}{4}$ ways that $6$ will be from the $60$ and $4$ will be from the $40$, thus the probability that the student will solve exactly $6$ questions is $$frac{binom{60}{6}binom{40}{4}}{binom{100}{10}}$$ assuming he solves every question he has prepared for. Similarly the probability that he has prepared for exactly $7$ questions is $$frac{binom{60}{7}binom{40}{3}}{binom{100}{10}} $$ and so on.
$endgroup$
– N. Shales
Mar 31 '17 at 15:20