Complicated venn diagram!

Question

I have been trying to work out the following Venn diagram question for the past 3 hours and I have not been able to work it out.

The main reason why I am struggling is because I dont have the value for the centre circle.

Please can someone provide input on how to solve the question.

Answer 1

Hint...put $x$ in the centre region, then work outwards filling each region with the amount in terms of $x$, so that, for example, $$n(B\cap M\cap L')=11-x$$

Answer 2

Let $A,B,C$ denote the events that someone has a bike, phone, and laptop, respectively.

For $(a)$ you can use the fact that $$P(A\cup\ B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A \cap C) - P(B \cap C) +P(A \cap B \cap C)$$

where $$P(A\cup\ B\cup C)=1-\frac{2}{32}$$

For $(b)$ note that

$$\begin{align*} P(A\cap B \cap C \mid (A\cap B) \cup (A \cap C) \cup (B\cap C)) &=\frac{P((A\cap B \cap C) \cap ((A\cap B) \cup (A \cap C) \cup (B\cap C)))}{P((A\cap B) \cup (A \cap C) \cup (B\cap C))}\\\\ &=\frac{P((A\cap B \cap C)}{P((A\cap B) \cup (A \cap C) \cup (B\cap C))}\\\\ \end{align*}$$

Answer 3

A venn diagram can help keep track of information in a visual format for three or fewer sets, but if and when you are asked to complete a similar problem where there are four or more sets, venn diagrams can become far too complicated or even impossible to draw on paper. For that reason, I suggest looking at a purely pen-and-paper approach that doesn't use such a diagram which doesn't have the same limitations.

Let $B$ represent the set of students with a bike, $P$ who have a phone, and $C$ who have a computer. Let $\Omega$ represent the set of all students in the class.

We are told the following information:

• The total number of students is $|\Omega|=32$

• The number of students who have a bike (among possibly other items) is $|B|=19$

• The number of students who have a phone (among possibly other items) is $|P|=21$

• The number of students who have a computer (among possibly other items) is $|C|=16$

• The number of students who have a bike and a phone (among possibly other items) is $|B\cap P|=11$

• The number of students who have a phone and a computer (among possibly other items) is $|P\cap C|=12$

• The number of students who have a bike and a computer (among possibly other items) is $|B\cap C|=6$

• The number of students who don't have any of the mentioned items is $|\Omega\setminus(B\cup P\cup C)|=2$

From the final piece of information, we learn that the number of students with at least one of the mentioned items is $|B\cup P\cup C|=|\Omega|-|\Omega\setminus(B\cup P\cup C)|=32-2=30$

Now... applying inclusion-exclusion principle, we know that

$$|B\cup P\cup C|=|B|+|P|+|C|-|B\cap P|-|B\cap C|-|P\cap C|+|B\cap P\cap C|$$

We know all of these pieces of information except the final one, $|B\cap P\cap C|$, the number of students who have all of a phone, bike, and computer, and this is precisely the number we are asked to calculate in the problem, so by plugging in the appropriate numbers we can find it.

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It is unclear exactly how you got stuck since you did not share any of your own efforts, but a common mistake would be that you interpreted certain phrases incorrectly. For example, the phrase "11 students have both a phone and a bike" means that these eleven students have a phone and a bike and some of them might have a computer too. I have seen several people assume that the $11$ students being referred to must all not have a computer and that any students with a phone, bike, and computer must be referred to elsewhere separately and aren't counted within those $11$ mentioned.