Answer to Question #98222 in Statistics and Probability for C

Question

Please explain to me how to draw Venn diagrams,for example when there are 3 choices like breakfast,lunch and dinner.Some eat breakfast only while some take both while some take all of them.Sometimes the values are not given and we have to work out the value of x.Sometimes the middle circle gets for example something minus x while another circle gets something plus x.Please show me an example like this and how I should draw and do this.Please explain to me,Thanks.I don’t know how to do it at all.

Answer 1

When you have elements to place in three sets, work from the inside out.

Step 1: Find the elements that are common to all three sets and place in region V.

Step 2: Find the elements for region II. Find the elements in $A\cap B.$ The elements in this set belong in regions II and V. Place the elements in the set that are not listed in region V in region II. The elements in regions IV and VI are found in a similar manner.

Step 3: Determine the elements to be placed in region I by determining the elements in set A that are not in regions II, IV, and V. The elements in regions III and VII are found in a similar manner.

Step 4: Determine the elements to be placed in region VIII by finding the elements in the universal set that are

not in regions I through VII.

$$Total=n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-$$ $$-n(B\cap C)+n(A\cap B\cap C)+n(No\ set)$$

$$Total=n(No\ set)+n(exactly\ one\ set)+$$ $$+n(exactly\ two\ sets)+n(exactly\ three\ sets)+$$

$Total=n(No\ set)+n(At\ least\ one\ set),$

$n(At\ least\ one \ set)=n(A)+n(B)+n(C)-$

$-n(A\cap B)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C),$

$n(A\cap B)$ includes the elements which are in both $A$ and $B$ and it also includes elements which are in $A,B$ and $C.$ Because of this we should remove $n(A\cap B\cap C)$ from $n(A\cap B)$ to get $n(A \ and\ B\ only).$

Similarly we get $n(A \ and\ C\ only)$ and $n(B \ and\ C\ only).$ So adding all these three give us number of elements in exactly 2 sets.

$n(Exactly\ two\ sets)=n(A\cap B)+n(A\cap C)+n(B \cap C)-$

$-3\cdot n(A\cap B\cap C)$

Now we can get $n(At\ least\ two\ sets).$ Here, we include the elements which are in all three sets once

$n(At\ least\ two\ sets)=n(Exactly\ two\ sets)+n(A\cap B\cap C)=$

$=n(A\cap B)+n(A\cap C)+n(B \cap C)-2\cdot n(A\cap B\cap C)$

$n(Exactly\ one\ set)=n(At\ least\ one \ set)-n(At\ least\ two\ sets)=$

$=n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(B\cap C)+$

$+n(A\cap B\cap C)-n(A\cap B)-n(A\cap C)-n(B \cap C)+$

$+2\cdot n(A\cap B\cap C)=n(A)+n(B)+n( C)-2\cdot n(A\cap B)-$

$-2\cdot n(A\cap C)-2\cdot n(B\cap C)+3\cdot n(A\cap B\cap C)$