In a town of 1000 households, 750 families each had a radio set, 421 houses each has a television set, while 120 houses had video sets each. If 70% of the families with video sets... In a town of 1000 households, 750 families each had a radio set, 421 houses each has a television set, while 120 houses had video sets each. If 70% of the families with video sets had no radio sets and every family with a video had a television set. Find: (a) Number of families with radio and television sets. (b) number of houses with all three. (c) number of houses with radio and television sets but without a video. (d) number of houses with television without radio sets Read more

Steps to Solve

1. Determine the number of families with video sets and no radio sets The problem states that 70% of families with video sets have no radio sets. There are 120 families with video sets. Therefore, the number of families with video sets and no radio sets is: $0.70 \times 120 = 84$

2. Determine the number of families with video and radio sets Since there are 120 families with video sets, and 84 of them do not have radio sets, the number of families with both video and radio sets is: $120 - 84 = 36$

3. Determine the number of families with radio and television sets Every family with a video set has a television set. Thus all 120 families with video sets have television sets. Of these, 36 also have radio sets (from step 2).

   Let $R$ be the event that a family has a radio set, and $T$ be the event that a family has a television set. We are given that $|R| = 750$ and $|T| = 421$. We also know that every family with a video set possesses a television set. Let $V$ denote the event that a family owns a video set. Thus, $V \subset T$, and $|V| = 120$. Furthermore, we already calculated the number of families that own a radio and video set, $|R \cap V| = 36$. Since $V \subset T$, $|V \cap R| = |T \cap V \cap R|$.

   We want to find $|R \cap T|$, the number of families with both radio and television sets. We know $|R| = 750$, $|T| = 421$. We also know that $|V| = 120$ and $V \subset T$, so all video set owners also have TV sets. 70% of video set owners do not have Radio sets i.e. 84 of the video set owners do not have radio sets. Therefore, 30% of video set owners do have radio sets i.e. 36 video set owners have radio sets. In set notation, $|V \cap R| = 36$.

   Since we cannot determine the exact overlap $|R \cap T|$ with the given information, we can only provide a range. The minimum number of families with both is $750 + 421 - 1000 = 171$ (using inclusion-exclusion). The maximum number of families with both is 421 (all TV owners also have radios). Let's assume that the question intended us to find a specific value and that it is relying on the information given about video sets somehow...

   However, we know every video set owner also has a television set. We also know that 36 video set owners also have radio sets. It is reasonable to assume that the video set information helps narrow in on the intersection of radio and television sets, although it does not strictly allow us to calculate $|R \cap T|$.

There's missing info to compute $|R \cap T|$ precisely, given the data. But based on all the statements, the best we can do is to assume that families owning video sets also owning radio sets contribute to this number $|R \cap T|$.

So, we can't compute the exact number of families with radio and television sets, given the provided information. We will have $|R \cap T| \ge |V \cap R| = 36$.

Based on question b requesting the "number of houses with all three" it seems that $|R \cap T|$ is necessary to work out this.

4. Determine the number of houses with all three sets Let $x$ be the number of houses with all three sets, i.e., $|R \cap T \cap V|$. Since every house with a video set also has a television set, all 120 houses with video sets also have television sets. We know from step 2 that 36 families have radio and video sets. Therefore, $x = |R \cap T \cap V| = |R \cap V| = 36$.

5. Determine the number of houses with radio and television sets but without a video set Let $y$ be the number of houses with radio and television sets but without a video set. This can be represented as $|R \cap T \cap V^c|$, where $V^c$ is the complement of $V$ (i.e., not having a video set). We know that $|R \cap T|$ represents the number of houses with both radio and television sets. We found that houses with all three is 36. We are looking for $|R \cap T \cap V^c| = |R \cap T| - |R \cap T \cap V|$, which is $|R \cap T| - 36$. Because we cannot compute $|R \cap T|$ exactly, we cannot compute $|R \cap T \cap V^c|$ exactly.

6. Determine the number of houses with television sets without radio sets The number of houses with television sets is $|T| = 421$. The number of houses with both television and radio sets is $|R \cap T|$. The number of houses with television sets but without radio sets is $|T| - |R \cap T| = 421 - |R \cap T|$. Again -- we cannot isolate $|R \cap T|$.

   Revisiting question a

   After considering the other questions, we will try estimating the quantity $|R \cap T|$. From step 3, $171 \le |R \cap T| \le 421$. Let us assume that 36 families owning the video sets and also owing radios all own television sets; let's assume these 36 are the only families that own radios and televisions. In that case the number of families with both radio and television sets equals 36. This does not lead to a result consistent with what we know about radio set ownership so let's ignore the assumption.

   We know that 84 families out of 120 have only video and television. I.e. television owners are split between solely televisions owners and television AND video owners. $421 - |R \cap T|$ also has to be greater than 84. 421 is all the TV owners, and $|R \cap T|$ is all the TV owners who also own radios. Meaning: 421 - radio AND television >= TV AND video, so subtract from both sides: 250 + 421 - 1000 <= radio AND television <= 421 - 84 171 <= radio AND television <= 337

7. Estimating final answers Since we know that the minimal overlap between $R$ and $T$, if we assume that $R \cap T = 171$, then Families owning 3 sets: 36 Families owning television without radio sets: $421 - 171 = 250$ Families owning television AND radio AND no video: $171 - 36 = 135$ Families owning television AND radio: 171