How Many Families Subscribe to Exactly One Newspaper?

[TheMathNoob]

Homework Statement

In a certain city, three newspapers A, B, and C are published. Suppose that 60 percent of the families in the city subscribe to newspaper A, 40 percent of the families subscribe to newspaper B, and 30 percent subscribe to newspaper C.Suppose also that 20 percent of the families subscribe to both A and B, 10 percent subscribe to both A and C, 20 percent subscribe to both B and C, and 5 percent subscribe to all three newspapers A, B, and C. what percentage of the families subscribe to exactly one of the newspapers.

Homework Equations

P(AUB)=P(A)+P(B)-P(BnC)
3. The Attempt at a Solution [/B]
I already find the percentage of the families that subscribe to at least 1 newspaper which resulted to be 85%. Doing the set thing, I realized that I have to find A B and C without the intersections like shading the whole area except for the complement and the intersections. I tried to subtract the 85% previously found from the the total sum of the intersections. Like this

A U B U C - A n C-A n B-B n C-A n B n C

But this is wrong because 1 person can be in one or more intersections.

There is solution in this website which I think that is wrong because they just add the intersections without realizing that one person can be in more than one intersections.

https://cseweb.ucsd.edu/classes/sp97/cse21/quizzes/q3solsph/q3solsph.html

 

[andrewkirk]

What is the question you are trying to answer? You have not stated it.

 

[andrewkirk]

Nevertheless, the following strategy will enable you to answer any question that is asked.

Draw a three-way symmetrical diagram of three intersecting circles. In the middle will be three leaf-shaped areas, each of which is the intersection of two circles. The three leaves intersect in a curvy triangular area that is the intersection of all three circles. You want to find the size - as a percentage - of every indivisible area (ie an area with no lines through it) in the diagram. The only such area of which you know the size at the beginning is the triangular bit, which is $A\cap B \cap C$.

You use that triangle as a starting point and work your way outwards, calculating the size of each other indivisible area in turn until you have all of them.

As a first step, why not work out the other part of the $A\cap B$ leaf, which is $A\cap B\cap \sim C$. You know $A\cap B$ and you know $A\cap B\cap C$. So $A\cap B\cap \sim C$ is ...??

 

[TheMathNoob]

Nevertheless, the following strategy will enable you to answer any question that is asked.

Draw a three-way symmetrical diagram of three intersecting circles. In the middle will be three leaf-shaped areas, each of which is the intersection of two circles. The three leaves intersect in a curvy triangular area that is the intersection of all three circles. You want to find the size - as a percentage - of every indivisible area (ie an area with no lines through it) in the diagram. The only such area of which you know the size at the beginning is the triangular bit, which is $A\cap B \cap C$.

You use that triangle as a starting point and work your way outwards, calculating the size of each other indivisible area in turn until you have all of them.

As a first step, why not work out the other part of the $A\cap B$ leaf, which is $A\cap B\cap \sim C$. You know $A\cap B$ and you know $A\cap B\cap C$. So $A\cap B\cap \sim C$ is ...??

Sorry, I already put the question.

 

[Ray Vickson]

Homework Statement

In a certain city, three newspapers A, B, and C are published. Suppose that 60 percent of the families in the city subscribe to newspaper A, 40 percent of the families subscribe to newspaper B, and 30 percent subscribe to newspaper C.Suppose also that 20 percent of the families subscribe to both A and B, 10 percent subscribe to both A and C, 20 percent subscribe to both B and C, and 5 percent subscribe to all three newspapers A, B, and C.

Homework Equations

P(AUB)=P(A)+P(B)-P(BnC)
3. The Attempt at a Solution [/B]
I already find the percentage of the families that subscribe to at least 1 newspaper which resulted to be 85%. Doing the set thing, I realized that I have to find A B and C without the intersections like shading the whole area except for the complement and the intersections. I tried to subtract the 85% previously found from the the total sum of the intersections. Like this

A U B U C - A n C-A n B-B n C-A n B n C

But this is wrong because 1 person can be in one or more intersections.

While you do not state it, I guess you want to know $P(A \cup B \cup C)$.

Anyway, you can get a correct approach by applying what you know, piece-by-piece. For any two sets $U, V$, I will use the notation $U V$ instead of $U \cap V$, just to simply writing.

You know already that $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 E_2)$ for any $E_1, E_2$. Apply this to $P(A \cup B \cup C)$ by setting $E_1 = A, E_2 = B \cup C$. You get $P(A \cup B \cup C) = P(A) + P(B \cup C) - P[A(B\cup C)]$. Now apply it again to $E_1 = B, E_2 = C$, and then to $E_1 = AB, E_2 = AC$. I will leave it to you to obtain the complete expression for $P(A \cup B \cup C)$ in terms of the known quantities you were given.

 

[TheMathNoob]

While you do not state it, I guess you want to know $P(A \cup B \cup C)$.

Anyway, you can get a correct approach by applying what you know, piece-by-piece. For any two sets $U, V$, I will use the notation $U V$ instead of $U \cap V$, just to simply writing.

You know already that $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 E_2)$ for any $E_1, E_2$. Apply this to $P(A \cup B \cup C)$ by setting $E_1 = A, E_2 = B \cup C$. You get $P(A \cup B \cup C) = P(A) + P(B \cup C) - P[A(B\cup C)]$. Now apply it again to $E_1 = B, E_2 = C$, and then to $E_1 = AB, E_2 = AC$.

I will leave it to you to write out the whole expression for $P(A \cup B \cup C)$.

Hi Ray, I already stated the question sorry

 

[Ray Vickson]

Hi Ray, I already stated the question sorry

Sorry: if you want $P\{ \text{exactly one paper} \} = P(A \, \text{only, or} \: B \,\text{only, or} \, C \: \text{only})$ then it is a bit more complex. You can write the desired event as
\{ \text{exactly one of } \, A, B, C \} = A' \cup B' \cup C',
where
A' = A - AB - AC, \: B' = B - BA - BC, \; C' = C - CA - CB
Notice that the sets $A', B', C'$ are disjoint, so the probability you want is
\text{answer} = P(A') + P(B') + P(C').

If you look at a Venn diagram you should see fairly easily how to get $P(A')$ or $P(B')$ or $P(C')$ in terms of the givens $P(A), P(B), P(C), P(AB), P(AC), P(BC)$ and $P(ABC)$.

 

[TheMathNoob]

Sorry: if you want $P\{ \text{exactly one paper} \} = P(A \, \text{only, or} \: B \,\text{only, or} \, C \: \text{only})$ then it is a bit more complex. You can write the desired event as
\{ \text{exactly one of } \, A, B, C \} = A' \cup B' \cup C',
where
A' = A - AB - AC, \: B' = B - BA - BC, \; C' = C - CA - CB
Notice that the sets $A', B', C'$ are disjoint, so the probability you want is
\text{answer} = P(A') + P(B') + P(C').

If you look at a Venn diagram you should see fairly easily how to get $P(A')$ or $P(B')$ or $P(C')$ in terms of the givens $P(A), P(B), P(C), P(AB), P(AC), P(BC)$ and $P(ABC)$.

what is ABC, BC?

 

[TheMathNoob]

Sorry: if you want $P\{ \text{exactly one paper} \} = P(A \, \text{only, or} \: B \,\text{only, or} \, C \: \text{only})$ then it is a bit more complex. You can write the desired event as
\{ \text{exactly one of } \, A, B, C \} = A' \cup B' \cup C',
where
A' = A - AB - AC, \: B' = B - BA - BC, \; C' = C - CA - CB
Notice that the sets $A', B', C'$ are disjoint, so the probability you want is
\text{answer} = P(A') + P(B') + P(C').

If you look at a Venn diagram you should see fairly easily how to get $P(A')$ or $P(B')$ or $P(C')$ in terms of the givens $P(A), P(B), P(C), P(AB), P(AC), P(BC)$ and $P(ABC)$.

I already did it. It was very challenging to me. The hard part was to figure out that 5% which was the percentage of people who read the three newspapers was in every intersection of the sets.

 

[Ray Vickson]

Homework Statement

In a certain city, three newspapers A, B, and C are published. Suppose that 60 percent of the families in the city subscribe to newspaper A, 40 percent of the families subscribe to newspaper B, and 30 percent subscribe to newspaper C.Suppose also that 20 percent of the families subscribe to both A and B, 10 percent subscribe to both A and C, 20 percent subscribe to both B and C, and 5 percent subscribe to all three newspapers A, B, and C. what percentage of the families subscribe to exactly one of the newspapers.

Homework Equations

P(AUB)=P(A)+P(B)-P(BnC)
3. The Attempt at a Solution [/B]
I already find the percentage of the families that subscribe to at least 1 newspaper which resulted to be 85%. Doing the set thing, I realized that I have to find A B and C without the intersections like shading the whole area except for the complement and the intersections. I tried to subtract the 85% previously found from the the total sum of the intersections. Like this

A U B U C - A n C-A n B-B n C-A n B n C

But this is wrong because 1 person can be in one or more intersections.

There is solution in this website which I think that is wrong because they just add the intersections without realizing that one person can be in more than one intersections.

https://cseweb.ucsd.edu/classes/sp97/cse21/quizzes/q3solsph/q3solsph.html

You are right: the website's answer is wrong, for exactly the reasons you give.