Venn diagram word problem
In a group of 100, 50 are having an umbrella, 60 have a hat and 80 have sunglasses. 70 are such that they don’t have both – an umbrella and a hat. Similarly 50 are such that they do not have both hat and sunglasses and 60 do not have both umbrellas and sunglasses. If there are 5 who do not own any of the three items, find the number of people who own all three items.
I couldn't solve the question using 2 equations
those who wear exactly one of the three -n (A ) + n (B ) + n (C ) - 2 (n (A ∩ B ) + n (B ∩ C ) + n (A ∩ C )) + 3 × n (A ∩ B ∩ C )
and the general formula to find the union -
n (A ) + n (B ) + n (C ) - n (A ∩ B ) - n (B ∩ C ) - n (A ∩ C ) .
I took the sum of 70 , 50 , 60 as the sum of those who wear exactly one of the three
and 95 as the union
elementary-set-theory
asked Nov 26 '18 at 7:29
yash modi
1
add a comment |
In a group of 100, 50 are having an umbrella, 60 have a hat and 80 have sunglasses. 70 are such that they don’t have both – an umbrella and a hat. Similarly 50 are such that they do not have both hat and sunglasses and 60 do not have both umbrellas and sunglasses. If there are 5 who do not own any of the three items, find the number of people who own all three items.
I couldn't solve the question using 2 equations
those who wear exactly one of the three -n (A ) + n (B ) + n (C ) - 2 (n (A ∩ B ) + n (B ∩ C ) + n (A ∩ C )) + 3 × n (A ∩ B ∩ C )
and the general formula to find the union -
n (A ) + n (B ) + n (C ) - n (A ∩ B ) - n (B ∩ C ) - n (A ∩ C ) .
I took the sum of 70 , 50 , 60 as the sum of those who wear exactly one of the three
and 95 as the union
elementary-set-theory
asked Nov 26 '18 at 7:29
yash modi
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 26 '18 at 7:30
I would always draw a diagram for a question like this. I find it rather easy to get confused.
– Mark Bennet
Nov 26 '18 at 8:03
add a comment |
In a group of 100, 50 are having an umbrella, 60 have a hat and 80 have sunglasses. 70 are such that they don’t have both – an umbrella and a hat. Similarly 50 are such that they do not have both hat and sunglasses and 60 do not have both umbrellas and sunglasses. If there are 5 who do not own any of the three items, find the number of people who own all three items.
I couldn't solve the question using 2 equations
those who wear exactly one of the three -n (A ) + n (B ) + n (C ) - 2 (n (A ∩ B ) + n (B ∩ C ) + n (A ∩ C )) + 3 × n (A ∩ B ∩ C )
and the general formula to find the union -
n (A ) + n (B ) + n (C ) - n (A ∩ B ) - n (B ∩ C ) - n (A ∩ C ) .
I took the sum of 70 , 50 , 60 as the sum of those who wear exactly one of the three
and 95 as the union
elementary-set-theory
asked Nov 26 '18 at 7:29
yash modi
1
In a group of 100, 50 are having an umbrella, 60 have a hat and 80 have sunglasses. 70 are such that they don’t have both – an umbrella and a hat. Similarly 50 are such that they do not have both hat and sunglasses and 60 do not have both umbrellas and sunglasses. If there are 5 who do not own any of the three items, find the number of people who own all three items.
I couldn't solve the question using 2 equations
those who wear exactly one of the three -n (A ) + n (B ) + n (C ) - 2 (n (A ∩ B ) + n (B ∩ C ) + n (A ∩ C )) + 3 × n (A ∩ B ∩ C )
and the general formula to find the union -
n (A ) + n (B ) + n (C ) - n (A ∩ B ) - n (B ∩ C ) - n (A ∩ C ) .
I took the sum of 70 , 50 , 60 as the sum of those who wear exactly one of the three
and 95 as the union
elementary-set-theory
elementary-set-theory
asked Nov 26 '18 at 7:29
yash modi
1
asked Nov 26 '18 at 7:29
yash modi
1
asked Nov 26 '18 at 7:29
yash modi
1
asked Nov 26 '18 at 7:29
yash modi
1
asked Nov 26 '18 at 7:29
yash modi
1
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 26 '18 at 7:30
I would always draw a diagram for a question like this. I find it rather easy to get confused.
– Mark Bennet
Nov 26 '18 at 8:03
add a comment |
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 26 '18 at 7:30
I would always draw a diagram for a question like this. I find it rather easy to get confused.
– Mark Bennet
Nov 26 '18 at 8:03
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 26 '18 at 7:30
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 26 '18 at 7:30
I would always draw a diagram for a question like this. I find it rather easy to get confused.
– Mark Bennet
Nov 26 '18 at 8:03
I would always draw a diagram for a question like this. I find it rather easy to get confused.
– Mark Bennet
Nov 26 '18 at 8:03
add a comment |
1 Answer
1
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Let $A$ be the set of people who own an umbrella, $B$ be the set of people who own a hat, and $C$ be the set of people who own sunglasses. We want to compute the number of people who own all three items, which is equal to $|A cap B cap C|.$
The Principle of Inclusion-Exclusion asserts
$$|A cup B cup C| = |A| + |B| + |C| - |A cap B| - |B cap C| - |A cap C| + |Acap Bcap C|. $$
Here, $|A| = 50, |B| = 60, $ and $|C| = 80$. Also, we know there are $70$ people who are not in the set $A cap B$, which means that there are $100 - 70 = 30$ people in this set. Thus, $|A cap B| = 30$. Using similar reasoning, we can find $|B cap C| = 50$ and $|A cap C| = 40$. Also, we know $|A cup B cup C| = 5$ for the reason you described. Therefore,
$$|A| + |B| + |C| - |A cap B| - |B cap C| - |A cap C| + |Acap Bcap C| $$
$$5= 50 + 60 + 80 - 30 - 50 - 40 - |A cap B cap C| $$
$$implies |A cap B cap C| = boxed{65} $$
answered Nov 26 '18 at 7:38
Ekesh
5326
Well the answer given for this particular problem is 25
– yash modi
Nov 27 '18 at 6:37
add a comment |
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1 Answer
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Let $A$ be the set of people who own an umbrella, $B$ be the set of people who own a hat, and $C$ be the set of people who own sunglasses. We want to compute the number of people who own all three items, which is equal to $|A cap B cap C|.$
The Principle of Inclusion-Exclusion asserts
$$|A cup B cup C| = |A| + |B| + |C| - |A cap B| - |B cap C| - |A cap C| + |Acap Bcap C|. $$
Here, $|A| = 50, |B| = 60, $ and $|C| = 80$. Also, we know there are $70$ people who are not in the set $A cap B$, which means that there are $100 - 70 = 30$ people in this set. Thus, $|A cap B| = 30$. Using similar reasoning, we can find $|B cap C| = 50$ and $|A cap C| = 40$. Also, we know $|A cup B cup C| = 5$ for the reason you described. Therefore,
$$|A| + |B| + |C| - |A cap B| - |B cap C| - |A cap C| + |Acap Bcap C| $$
$$5= 50 + 60 + 80 - 30 - 50 - 40 - |A cap B cap C| $$
$$implies |A cap B cap C| = boxed{65} $$
answered Nov 26 '18 at 7:38
Ekesh
5326
Well the answer given for this particular problem is 25
– yash modi
Nov 27 '18 at 6:37
add a comment |
Let $A$ be the set of people who own an umbrella, $B$ be the set of people who own a hat, and $C$ be the set of people who own sunglasses. We want to compute the number of people who own all three items, which is equal to $|A cap B cap C|.$
The Principle of Inclusion-Exclusion asserts
$$|A cup B cup C| = |A| + |B| + |C| - |A cap B| - |B cap C| - |A cap C| + |Acap Bcap C|. $$
Here, $|A| = 50, |B| = 60, $ and $|C| = 80$. Also, we know there are $70$ people who are not in the set $A cap B$, which means that there are $100 - 70 = 30$ people in this set. Thus, $|A cap B| = 30$. Using similar reasoning, we can find $|B cap C| = 50$ and $|A cap C| = 40$. Also, we know $|A cup B cup C| = 5$ for the reason you described. Therefore,
$$|A| + |B| + |C| - |A cap B| - |B cap C| - |A cap C| + |Acap Bcap C| $$
$$5= 50 + 60 + 80 - 30 - 50 - 40 - |A cap B cap C| $$
$$implies |A cap B cap C| = boxed{65} $$
answered Nov 26 '18 at 7:38
Ekesh
5326
Well the answer given for this particular problem is 25
– yash modi
Nov 27 '18 at 6:37
add a comment |
Let $A$ be the set of people who own an umbrella, $B$ be the set of people who own a hat, and $C$ be the set of people who own sunglasses. We want to compute the number of people who own all three items, which is equal to $|A cap B cap C|.$
The Principle of Inclusion-Exclusion asserts
$$|A cup B cup C| = |A| + |B| + |C| - |A cap B| - |B cap C| - |A cap C| + |Acap Bcap C|. $$
Here, $|A| = 50, |B| = 60, $ and $|C| = 80$. Also, we know there are $70$ people who are not in the set $A cap B$, which means that there are $100 - 70 = 30$ people in this set. Thus, $|A cap B| = 30$. Using similar reasoning, we can find $|B cap C| = 50$ and $|A cap C| = 40$. Also, we know $|A cup B cup C| = 5$ for the reason you described. Therefore,
$$|A| + |B| + |C| - |A cap B| - |B cap C| - |A cap C| + |Acap Bcap C| $$
$$5= 50 + 60 + 80 - 30 - 50 - 40 - |A cap B cap C| $$
$$implies |A cap B cap C| = boxed{65} $$
answered Nov 26 '18 at 7:38
Ekesh
5326
Let $A$ be the set of people who own an umbrella, $B$ be the set of people who own a hat, and $C$ be the set of people who own sunglasses. We want to compute the number of people who own all three items, which is equal to $|A cap B cap C|.$
The Principle of Inclusion-Exclusion asserts
$$|A cup B cup C| = |A| + |B| + |C| - |A cap B| - |B cap C| - |A cap C| + |Acap Bcap C|. $$
Here, $|A| = 50, |B| = 60, $ and $|C| = 80$. Also, we know there are $70$ people who are not in the set $A cap B$, which means that there are $100 - 70 = 30$ people in this set. Thus, $|A cap B| = 30$. Using similar reasoning, we can find $|B cap C| = 50$ and $|A cap C| = 40$. Also, we know $|A cup B cup C| = 5$ for the reason you described. Therefore,
$$|A| + |B| + |C| - |A cap B| - |B cap C| - |A cap C| + |Acap Bcap C| $$
$$5= 50 + 60 + 80 - 30 - 50 - 40 - |A cap B cap C| $$
$$implies |A cap B cap C| = boxed{65} $$
answered Nov 26 '18 at 7:38
Ekesh
5326
answered Nov 26 '18 at 7:38
Ekesh
5326
answered Nov 26 '18 at 7:38
Ekesh
5326
answered Nov 26 '18 at 7:38
Ekesh
5326
5326
Well the answer given for this particular problem is 25
– yash modi
Nov 27 '18 at 6:37
add a comment |
Well the answer given for this particular problem is 25
– yash modi
Nov 27 '18 at 6:37
Well the answer given for this particular problem is 25
– yash modi
Nov 27 '18 at 6:37
Well the answer given for this particular problem is 25
– yash modi
Nov 27 '18 at 6:37
add a comment |
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 26 '18 at 7:30
I would always draw a diagram for a question like this. I find it rather easy to get confused.
– Mark Bennet
Nov 26 '18 at 8:03