Mixing
How to find the intersection part in a Venn diagram?
$begingroup$
My problem: There are $20$ students in a class. $13$ of them study chemistry and $16$ of them study physics. $3$ of them do neither.
My workings:
$13 + 16 = 29$, but there are only $20$ people in the class, that means some people have to do both right? But how do I determine $C cap P$? So, I said $20 -3 =17$, then $C cup P$ must have a total of $17$ people. But then I am stuck.
Can someone please visually represent this? Mathematically showing how to find $C cap P$ is also fine.
probability
edited Feb 1 at 16:10
jvdhooft
5,65961641
asked Feb 1 at 13:03
Fred WeasleyFred Weasley
16910
$endgroup$
add a comment |
$begingroup$
My problem: There are $20$ students in a class. $13$ of them study chemistry and $16$ of them study physics. $3$ of them do neither.
My workings:
$13 + 16 = 29$, but there are only $20$ people in the class, that means some people have to do both right? But how do I determine $C cap P$? So, I said $20 -3 =17$, then $C cup P$ must have a total of $17$ people. But then I am stuck.
Can someone please visually represent this? Mathematically showing how to find $C cap P$ is also fine.
probability
edited Feb 1 at 16:10
jvdhooft
5,65961641
asked Feb 1 at 13:03
Fred WeasleyFred Weasley
16910
$endgroup$
add a comment |
$begingroup$
My problem: There are $20$ students in a class. $13$ of them study chemistry and $16$ of them study physics. $3$ of them do neither.
My workings:
$13 + 16 = 29$, but there are only $20$ people in the class, that means some people have to do both right? But how do I determine $C cap P$? So, I said $20 -3 =17$, then $C cup P$ must have a total of $17$ people. But then I am stuck.
Can someone please visually represent this? Mathematically showing how to find $C cap P$ is also fine.
probability
edited Feb 1 at 16:10
jvdhooft
5,65961641
asked Feb 1 at 13:03
Fred WeasleyFred Weasley
16910
$endgroup$
My problem: There are $20$ students in a class. $13$ of them study chemistry and $16$ of them study physics. $3$ of them do neither.
My workings:
$13 + 16 = 29$, but there are only $20$ people in the class, that means some people have to do both right? But how do I determine $C cap P$? So, I said $20 -3 =17$, then $C cup P$ must have a total of $17$ people. But then I am stuck.
Can someone please visually represent this? Mathematically showing how to find $C cap P$ is also fine.
probability
probability
edited Feb 1 at 16:10
jvdhooft
5,65961641
asked Feb 1 at 13:03
Fred WeasleyFred Weasley
16910
edited Feb 1 at 16:10
jvdhooft
5,65961641
asked Feb 1 at 13:03
Fred WeasleyFred Weasley
16910
edited Feb 1 at 16:10
jvdhooft
5,65961641
edited Feb 1 at 16:10
jvdhooft
5,65961641
edited Feb 1 at 16:10
jvdhooft
5,65961641
5,65961641
asked Feb 1 at 13:03
Fred WeasleyFred Weasley
16910
asked Feb 1 at 13:03
Fred WeasleyFred Weasley
16910
asked Feb 1 at 13:03
Fred WeasleyFred Weasley
16910
16910
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
As you correctly observed, $17$ study either chemistry or physics. As $16$ study physics only $1$ student studies only chemistry. Similarly, $4$ students study only physics. This results in $12$ students studying both.
answered Feb 1 at 13:11
KlausKlaus
2,955214
$endgroup$
add a comment |
$begingroup$
You have a big set of Students $S$ and you have two subsets $A,B$ of students who study chemistry or physics respectively. You know that $#(Acup B)=20-3=17$.
Also note that $Acap B = (A^ccup B^c)^c$ where $c$ denotes the complement. But you do know how many people do NOT study chemistry or physics.
Do you know how to continue from here?
answered Feb 1 at 13:19
Rene RecktenwaldRene Recktenwald
1808
$endgroup$
add a comment |
$begingroup$
If $C$ is the set containing all chemistry students and $P$ is the set containing all physics students, we have:
$$|C| + |P| - |C cap P| = 20 - 3 iff 13 + 16 - |C cap P| = 17$$
$$iff |C cap P| = 29 - 17 = 12$$
We thus find $12$ students studying both topics, $13 - 12 = 1$ studying only chemistry and $16 - 12 = 4$ studying only physics. The Venn diagram looks as follows:
answered Feb 1 at 13:15
jvdhooftjvdhooft
5,65961641
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As you correctly observed, $17$ study either chemistry or physics. As $16$ study physics only $1$ student studies only chemistry. Similarly, $4$ students study only physics. This results in $12$ students studying both.
answered Feb 1 at 13:11
KlausKlaus
2,955214
$endgroup$
add a comment |
$begingroup$
As you correctly observed, $17$ study either chemistry or physics. As $16$ study physics only $1$ student studies only chemistry. Similarly, $4$ students study only physics. This results in $12$ students studying both.
answered Feb 1 at 13:11
KlausKlaus
2,955214
$endgroup$
add a comment |
$begingroup$
As you correctly observed, $17$ study either chemistry or physics. As $16$ study physics only $1$ student studies only chemistry. Similarly, $4$ students study only physics. This results in $12$ students studying both.
answered Feb 1 at 13:11
KlausKlaus
2,955214
$endgroup$
As you correctly observed, $17$ study either chemistry or physics. As $16$ study physics only $1$ student studies only chemistry. Similarly, $4$ students study only physics. This results in $12$ students studying both.
answered Feb 1 at 13:11
KlausKlaus
2,955214
answered Feb 1 at 13:11
KlausKlaus
2,955214
answered Feb 1 at 13:11
KlausKlaus
2,955214
answered Feb 1 at 13:11
KlausKlaus
2,955214
2,955214
add a comment |
add a comment |
$begingroup$
You have a big set of Students $S$ and you have two subsets $A,B$ of students who study chemistry or physics respectively. You know that $#(Acup B)=20-3=17$.
Also note that $Acap B = (A^ccup B^c)^c$ where $c$ denotes the complement. But you do know how many people do NOT study chemistry or physics.
Do you know how to continue from here?
answered Feb 1 at 13:19
Rene RecktenwaldRene Recktenwald
1808
$endgroup$
add a comment |
$begingroup$
You have a big set of Students $S$ and you have two subsets $A,B$ of students who study chemistry or physics respectively. You know that $#(Acup B)=20-3=17$.
Also note that $Acap B = (A^ccup B^c)^c$ where $c$ denotes the complement. But you do know how many people do NOT study chemistry or physics.
Do you know how to continue from here?
answered Feb 1 at 13:19
Rene RecktenwaldRene Recktenwald
1808
$endgroup$
add a comment |
$begingroup$
You have a big set of Students $S$ and you have two subsets $A,B$ of students who study chemistry or physics respectively. You know that $#(Acup B)=20-3=17$.
Also note that $Acap B = (A^ccup B^c)^c$ where $c$ denotes the complement. But you do know how many people do NOT study chemistry or physics.
Do you know how to continue from here?
answered Feb 1 at 13:19
Rene RecktenwaldRene Recktenwald
1808
$endgroup$
You have a big set of Students $S$ and you have two subsets $A,B$ of students who study chemistry or physics respectively. You know that $#(Acup B)=20-3=17$.
Also note that $Acap B = (A^ccup B^c)^c$ where $c$ denotes the complement. But you do know how many people do NOT study chemistry or physics.
Do you know how to continue from here?
answered Feb 1 at 13:19
Rene RecktenwaldRene Recktenwald
1808
answered Feb 1 at 13:19
Rene RecktenwaldRene Recktenwald
1808
answered Feb 1 at 13:19
Rene RecktenwaldRene Recktenwald
1808
answered Feb 1 at 13:19
Rene RecktenwaldRene Recktenwald
1808
1808
add a comment |
add a comment |
$begingroup$
If $C$ is the set containing all chemistry students and $P$ is the set containing all physics students, we have:
$$|C| + |P| - |C cap P| = 20 - 3 iff 13 + 16 - |C cap P| = 17$$
$$iff |C cap P| = 29 - 17 = 12$$
We thus find $12$ students studying both topics, $13 - 12 = 1$ studying only chemistry and $16 - 12 = 4$ studying only physics. The Venn diagram looks as follows:
answered Feb 1 at 13:15
jvdhooftjvdhooft
5,65961641
$endgroup$
add a comment |
$begingroup$
If $C$ is the set containing all chemistry students and $P$ is the set containing all physics students, we have:
$$|C| + |P| - |C cap P| = 20 - 3 iff 13 + 16 - |C cap P| = 17$$
$$iff |C cap P| = 29 - 17 = 12$$
We thus find $12$ students studying both topics, $13 - 12 = 1$ studying only chemistry and $16 - 12 = 4$ studying only physics. The Venn diagram looks as follows:
answered Feb 1 at 13:15
jvdhooftjvdhooft
5,65961641
$endgroup$
add a comment |
$begingroup$
If $C$ is the set containing all chemistry students and $P$ is the set containing all physics students, we have:
$$|C| + |P| - |C cap P| = 20 - 3 iff 13 + 16 - |C cap P| = 17$$
$$iff |C cap P| = 29 - 17 = 12$$
We thus find $12$ students studying both topics, $13 - 12 = 1$ studying only chemistry and $16 - 12 = 4$ studying only physics. The Venn diagram looks as follows:
answered Feb 1 at 13:15
jvdhooftjvdhooft
5,65961641
$endgroup$
If $C$ is the set containing all chemistry students and $P$ is the set containing all physics students, we have:
$$|C| + |P| - |C cap P| = 20 - 3 iff 13 + 16 - |C cap P| = 17$$
$$iff |C cap P| = 29 - 17 = 12$$
We thus find $12$ students studying both topics, $13 - 12 = 1$ studying only chemistry and $16 - 12 = 4$ studying only physics. The Venn diagram looks as follows:
answered Feb 1 at 13:15
jvdhooftjvdhooft
5,65961641
edited Feb 1 at 13:28
answered Feb 1 at 13:15
jvdhooftjvdhooft
5,65961641
answered Feb 1 at 13:15
jvdhooftjvdhooft
5,65961641
answered Feb 1 at 13:15
jvdhooftjvdhooft
5,65961641
5,65961641
add a comment |
add a comment |
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