Question

solve the problem.

1. results of a survey of fifty students indicate that 30 like red jelly beans, 29 like green jelly beans, and 17 like both red and green jelly beans. how many of the students surveyed like only green jelly beans?
2. a local television station sends out questionnaires to determine if viewers would rather see a documentary, an interview show, or reruns of a game show. there were 600 responses with the following results:

180 were interested in an interview show and a documentary, but not reruns.
24 were interested in an interview show and reruns but not a documentary
84 were interested in reruns but not an interview show.
144 were interested in an interview show but not a documentary.
60 were interested in a documentary and reruns.
36 were interested in an interview show and reruns.
48 were interested in none of the three.
how many are interested in exactly one kind of show?

Explanation:

Step1: Find number of only - green jelly - bean likers

We know the formula for the number of elements in a set. Let \(R\) be the set of students who like red jelly - beans and \(G\) be the set of students who like green jelly - beans. The number of students who like green jelly - beans \(n(G)=29\), and the number of students who like both \(n(R\cap G)=17\). The number of students who like only green jelly - beans is \(n(G)-n(R\cap G)\).
\[29 - 17=12\]

Step2: Analyze the television - show problem

Let \(I\) be the set of viewers interested in an interview show, \(D\) be the set of viewers interested in a documentary, and \(R\) be the set of viewers interested in reruns.
We are given:
- Number interested in \(I\) and \(D\) but not \(R\): \(n(I\cap D\cap\overline{R}) = 180\)
- Number interested in \(I\) and \(R\) but not \(D\): \(n(I\cap\overline{D}\cap R)=24\)
- Number interested in \(R\) but not \(I\): \(n(\overline{I}\cap R)=84\)
- Number interested in \(I\) but not \(D\): \(n(I\cap\overline{D}) = 144\)
- Number interested in \(D\) and \(R\): \(n(D\cap R)=60\)
- Number interested in \(I\) and \(R\): \(n(I\cap R)=36\)
- Number interested in none of the three: \(48\)
First, find the number interested in only \(I\): \(n(I\cap\overline{D}\cap\overline{R})=n(I\cap\overline{D})-n(I\cap\overline{D}\cap R)=144 - 24=120\)
The number interested in only \(D\): Let \(x=n(D\cap\overline{I}\cap\overline{R})\).
We know that the total number of responses is \(N = 600\) and the number of non - interested is \(48\), so the number of interested in at least one is \(600 - 48=552\).
We can use the principle of inclusion - exclusion. But to find the number interested in exactly one kind of show, we sum the number interested in only \(I\), only \(D\), and only \(R\).
The number interested in only \(R\): \(n(\overline{I}\cap\overline{D}\cap R)=n(\overline{I}\cap R)-n(I\cap\overline{D}\cap R)=84 - 24 = 60\)
The number interested in exactly one kind of show is \(n(I\cap\overline{D}\cap\overline{R})+n(\overline{I}\cap D\cap\overline{R})+n(\overline{I}\cap\overline{D}\cap R)\)
We find \(n(\overline{I}\cap D\cap\overline{R})\) using the fact that \(n(D\cap R) = 60\) and \(n(I\cap D\cap R)\) (from \(n(I\cap R)\) and other intersections, \(n(I\cap D\cap R)=36 - 24=12\)), and then from the total number of interested in \(D\) related sets. But we can also calculate the number interested in exactly one as follows:
The number interested in only \(I\): \(144-24 = 120\)
The number interested in only \(R\): \(84 - 24=60\)
The number interested in only \(D\): We know that the sum of all non - zero intersections and non - interested should equal \(600\).
The number interested in exactly one kind of show is \(120+(n(D\cap\overline{I}\cap\overline{R}))+60\)
We calculate \(n(D\cap\overline{I}\cap\overline{R})\) as follows:
The sum of all given non - zero intersections and non - interested:
\(180+24 + 84+144+60+36+48=586\) (this is wrong approach above, let's start over for part 2)
Number interested in only \(I\): \(144 - 24=120\)
Number interested in only \(R\): \(84\)
Number interested in only \(D\):
The total number of non - non - interested (\(600 - 48 = 552\))
We know that \(n(I\cap D\cap\overline{R}) = 180\), \(n(I\cap\overline{D}\cap R)=24\), \(n(\overline{I}\cap D\cap R)=60 - 12 = 48\) (since \(n(I\cap R) = 36\) and \(n(I\cap\overline{D}\cap R)=24\), so \(n(I\cap D\cap R)=36 - 24 = 12\) and \(n(\overline{I}\cap D\cap R)=n(D\cap R)-n(I\cap D\cap R)\))
Let \(a\) be number interested in only \(I\), \(b\) be number interested in only \(D\), \(c\) be number interested in only \(R\)
\(a=144 - 24=120\), \(c = 84\)
We know that \(n(I\cap D\cap\overline{R})+n(I\cap\overline{D}\cap R)+n(\overline{I}\cap D\cap R)+a + b + c+48=600\)
We also know that \(n(I\cap D\cap\overline{R}) = 180\), \(n(I\cap\overline{D}\cap R)=24\), \(n(\overline{I}\cap D\cap R)=48\)
\(180+24 + 48+120 + b+84+48=600\)
\(b=600-(180 + 24+48+120+84+48)=96\)
The number interested in exactly one kind of show is \(120+96 + 84=300\)

Answer:

1. \(12\)
2. \(300\)