A certain town with a population of 100,000 has 3 newspapers: I, II, and III. The proportions of townspeople who read these papers are as follows:

I: 10 percent I and II: 8 percent I and II and III: 1 percent

II: 30 percent I and III: 2 percent

III: 5 percent II and III: 4 percent

(The list tells us, for instance, that 8000 people
read newspapers I and II.)

(a) Find the number of people who read only one
newspaper.

(b) How many people read at least two
newspapers?

(c) If I and III are morning papers and II is an
evening paper, how many people read at least
one morning paper plus an evening paper?

(d) How many people do not read any
newspapers?

(e) How many people read only one morning
paper and one evening paper?

probability

venn diagrams

Answered by Aaron 10 months ago

It is best to create a **Venn Diagram** for this type of problem,

**Find the number of people who read only one newspaper**

*People who only read newspaper I*

$$P(I \cap II^c \cap III^c)=P(I)-P(I \cap II)-P(I\cap III)+P(I\cap II\cap III)=10-8-2+1=1\%$$

This makes sense intuitively because we subtract the middle portion twice (I,II, and III), so we have to add it back in once.

Similarly for newspapers II and III we have,

$$P(I^c \cap II \cap III^c)=30-8-4+1 = 19\%$$

$$P(I^c \cap II^c \cap III) = 5-4-2+1 = 0\%$$

So the number of people who read only one newspaper is,

$1+19+0 = 20\%$

*20,000 people*

**How many people read at least two newspapers?**

This consists of all the middle portions, I and II, I and III, II and III, and I,II,III.

$$P(I \cap II \cap III^c) = P(I\cap II) - P(I \cap II \cap III) = 8 - 1 = 7\%$$

similarly, $$P(I \cap II^c \cap III) = 2 - 1 = 1\%$$ $$P(I^c \cap II \cap III) = 4 - 1 = 3\%$$

and hence people who read at least two newspapers totals,

$7+1+3+1 = 12\%$

*12,000 people*

**If I and III are morning papers and II is an evening paper, how many people read at least one morning paper plus an evening paper**

We did most of the work in **part (b)**, the answer is,

$P(I \cap II \cap III^c) + P(I \cap II \cap III) + P(I^c \cap II \cap III) = 7+1+3 = 11\% $

*11,000 people*

**How many people do not read any newspapers?**

The amount of people who do not read any newspapers are 100% minus the amount of all readers,

$$P(\text{do not read any newspapers})=100\%-P(I \cup II \cup III)$$

to find $P(I \cup II \cup III)$ use DeMorgan's theorem for three sets,

$$P(I \cup II \cup III) = P(I)+P(II)+P(III)-P(I \cap II) - P(I \cap III) - P(II \cap III) + P(I \cap II \cap III)$$

$$P(I \cup II \cup III) = 10 + 30 + 5 - 8 - 2 - 4 + 1 = 32\%$$

therefore,

$$P(\text{do not read any newspapers})=100\%-32\%=68\%$$

*68,000 people*

**How many people read only one morning paper and one evening paper?**

Again, most of the work is done in **part (b)**,

$P(I \cap II \cap III^c) + P(I^c \cap II \cap III) = 7+3 = 10\%$

*10,000 people*

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Asked: 10 months ago