Venn diagram

I don't to see the diagram but I do need to see the entire question post in readable English.
 
IMG_20201011_235941.jpgIMG_20201011_235951.jpg
What ive done
IMG_20201012_000019.jpg
I dont even know where im stuck maybe because i dont really understand the question statement.
I think the questions dont give enough information to solve it.
 
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Having strained my eyes and played cryptographer, it appears that

[MATH]n(P) = 165, \ n(Q) = 105,\ n(R) = 95, \\ n \left (P \bigcup Q \bigcup R \right ) = 255, \ \left (P \bigcap Q \right ) = 40, \\ \left (Q \bigcap R \right ) = 35, \ \left (P \bigcap R \right ) = 60,\\ \text {and } \left (P \bigcap Q \bigcap R\right ) = 10.[/MATH]I am not 100% positive about that because the English is not entirely free of ambiguity.

The question is to find

[MATH]n \left (P \bigcap Q \bigcap \neg R \right ).[/MATH]
To the OP: is that what you think it means?
 
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JeffM said:
Having strained my eyes and played cryptographer, it appears that

[MATH]n(P) = 165, \ n(Q) = 105,\ n(R) = 95, \\ n \left (P \bigcup Q \bigcup R \right ) = 255, \ \left (P \bigcap Q \right ) = 40, \\ \left (Q \bigcap R \right ) = 35, \ \left (P \bigcap R \right ) = 60,\\ \text {and } \left (P \bigcap Q \bigcap R\right ) = 10.[/MATH]I am not 100% positive about that because the English is not entirely free of ambiguity.

The question is to find

[MATH]n \left (P \bigcap Q \bigcap \neg R \right ).[/MATH]
To the OP: is that what you think it means?
Click to expand...
Yeah but i dont know how
 
@pka

pka taught this subject, and I hope he will give a more orderly explanation than I can, but I shall try to give a very common sense answer. I suggest, however, that you send a clear picture of the entire problem. Right now, I am not sure we understand it correctly. It would also help if you tell us a little bit about your level of education so we can give an explanation suitable for you

Let's start by thinking about groups that have NO MEMBERS in common, like the boys in a class and the girls in a class. How many are in the class? We simply add the number of boys in the class and the number of girls in the class. Easy. Any question?

OK. Here is a problem. In a class, 14 like chocolate ice cream best, 10 like vanilla ice cream best. But there are 31 students in the class. How is this possible? 14 + 10 = 24 < 31. There must be 7 students who like best something other than either chocolate or vanilla. If the sum of those in the groups we counted add up to LESS THAN the total, there must be groups we failed to count. Make sense?

All right. Now let's think about a slightly more complex problem. In a room, we have 26 diplomats. 13 diplomats served in Tokyo; 9 served in Brussels, and 6 served in neither city. 13 + 9 + 6 = 28 > 26. How is this possible? We must have overcounted somewhere. Well, we did. 2 diplomats must have served in both Tokyo and Brussels. We really must have

11 served only in Tokyo
7 served only in Brussels
2 served in both
6 served in neither

11 + 7 + 2 + 6 = 26.

We now see the importance of ensuring that two or more groups do not contain common members or else that an adjustment is made to account for common members.

We can get the correct answer in either one of two ways

13 + 9 + 6 - 2 = 26 = 11 + 7 + 2 + 6.

You with me?

Now I like to break things down into groups without common members ("disjoint sets").

Now, in your problem, how many prefer all three brands?

How many prefer P and Q?

Does that include those who like all three brands?

How many prefer just P and Q but not R?

How many prefer just P and R but not Q?

How many prefer just Q and R but not P?
 
Diagram 15 is a Venn diagram that showsthe result of a survey carried out by a company on a group of 255 people about their favourite handphones.IMG_20201012_234535.png
It is given that n(P) = 165,n(Q)=105 and n (Q) = 95. There are 40 people preferred both brand P and brand Q, 35 people preferred both brand Q and brand R, 60 people preferred both brand P and brand R and 10 people preferred all three brands of handphones.

How many people preferred brand P and brand Q but not brand R? 16025163918925149397999584568954.jpg

My answer :30
Is it correct? [MATH][/MATH]
 
vennnMHB.gif
This Venn diagram is based upon Jeff' reading of the data.
If correct then look diagram \(\#(P\cap Q\cap\neg R)=30\)
 
I do not view Venn diagrams as useful for solving a problem. They are useful for presenting an answer after you have found it.

Let's think about this problem AS I UNDERSTAND IT.

255 people were asked three questions. Whether they liked P brand. Whether they liked Q brand? And whether they liked R brand? Each person could say Y for yes or N for no to each question. Thus we have 2 * 2 * 2 = 8 possible groups, depending how a person answered all three questions. Of course one or more of the groups may have no one in it. But no group has in it a person it who is in another group.

YYY for P, Q, and R respectively. Number in group is 10.
YYN for P, Q, and R respectively. Number in group is b.
YNY for P, Q, and R respectively. Number in group is c.
YNN for P, Q, and R respectively. Number in group is d.
NYY for P, Q, and R respectively. Number in group is e.
NYN for P, Q, and R respectively. Number in group is f.
NNY for P, Q, and R respectively. Number in group is g.
NNN for P, Q, and R respectively. Number in group is h.

If we combine the YYY and the YYN groups, we have the group that likes P and Q. We are told how many are in that combined group, which is 40. So 10 + a = 40. Therefore, a = 30. Make sense?

If we combine the YYY and YNY groups, we have the group that likes P and R. We are told that combined group has 60. So 10 + b = 60. Thus b = 50.

If we combine the YYY, YYN, YNY, and YNN, we have the group that likes P. We are told that big group contains 165. So,
10 + a + b + c = 165 = 10 + 30 + 50 + c = 165, which means c = 165 - 90 = 75.

Can you finish confirming the numbers pka has given you in the Venn diagram? What does the 15 represent?

It is probable that pka knows more efficient methods to proceed, but this is simple logic and basic arithmetic. If you understand it, this kind of problem should be easy for you in the future.
 
I have question for ArifRaihan, why are you using a completely non-standard diagram?
Look at mine consisting three circles that create eight regions.
This is @ArifRaihan & @Jeff, lets agree that \(\neg X\) stands for not in \(X\).
And \(\#(X\cap Y)\) stands for "the number in the common part of \(X~\&~Y
venn3df.gif
Always start with \(\#(X\cap Y\cap Z)=d\) the center of the diagram.
Now \(\#(X\cap Y)=d+b\) so \(\#(X\cap Y\cap \neg Z)=b\)
\(\#(x)=a+b+d+f)\) and \(\#(X\cup Y\cup Z)=a+b+c+d+e+g\)
\(\#(\neg X\cap\neg Y\cap\neg Z)=h\)
 
@pka

If you look at my original post, you will see that I used the negation sign, but I really like your use of # rather than n. I shall do the same in the future.

I can see that starting in the most nested compartments may make Venn diagrams a better tool for solving problems, but you still need to know how many compartments are relevant. So I continue to think they are better for explaining a result than finding it. Of course I may be wrong; I was never taught how to use such diagrams as a tool. My method of defining all disjoint sets and adding them up may be very unsophisticated, but it is not hard to grasp conceptually.

Unfortunately, I suspect Arif is in grade school or early in high school. I doubt giving him a better notation will help him if he does not understand how to think through this type of problem first. Notation is wonderful if you get how it captures a certain kind logic, but until then it just seems to the student like extra work. I strongly suspect that is the case here.
 
Thank you for your help
I lack interest in this subject at school plus i got easily confused by the statement thats y im struggling with it.
When I see this question on exam I skip it to do other question first.
 
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