In a particular locality, 25 people like walking, 22 people like running, and 15 people like cycling. 11 people like both walking and running, 9 people like both running and cycling, and 7 people like both walking and cycling. 3 people like all three activities, while 1 do not like any of the activities. We need to: (a) Draw a Venn diagram to illustrate this information. (b) Find how many people are in this locality. (c) Find how many people like walking only. (d) Find how many people like two different activities only.

Probability and StatisticsVenn DiagramsSet TheoryInclusion-Exclusion PrincipleCombinatorics
2025/3/21

1. Problem Description

In a particular locality, 25 people like walking, 22 people like running, and 15 people like cycling. 11 people like both walking and running, 9 people like both running and cycling, and 7 people like both walking and cycling. 3 people like all three activities, while 1 do not like any of the activities. We need to:
(a) Draw a Venn diagram to illustrate this information.
(b) Find how many people are in this locality.
(c) Find how many people like walking only.
(d) Find how many people like two different activities only.

2. Solution Steps

(a) Venn Diagram:
Let W be the set of people who like walking, R be the set of people who like running, and C be the set of people who like cycling. We fill in the Venn diagram from the inside out.
* WRC=3W \cap R \cap C = 3
* WR=11W \cap R = 11, so WRC=113=8W \cap R \cap C' = 11 - 3 = 8
* RC=9R \cap C = 9, so RCW=93=6R \cap C \cap W' = 9 - 3 = 6
* WC=7W \cap C = 7, so WCR=73=4W \cap C \cap R' = 7 - 3 = 4
* W=25W = 25, so W only =25834=10W \text{ only } = 25 - 8 - 3 - 4 = 10
* R=22R = 22, so R only =22836=5R \text{ only } = 22 - 8 - 3 - 6 = 5
* C=15C = 15, so C only =15436=2C \text{ only } = 15 - 4 - 3 - 6 = 2
* Number of people who do not like any activity = 1
(b) Total number of people in the locality:
WRC|W \cup R \cup C| = (W only) + (R only) + (C only) + (W and R only) + (R and C only) + (W and C only) + (All three) + (None)
=10+5+2+8+6+4+3+1=39= 10 + 5 + 2 + 8 + 6 + 4 + 3 + 1 = 39
(c) Number of people who like walking only:
From the Venn diagram, the number of people who like walking only is 1010.
(d) Number of people who like two different activities only:
This is the sum of the number of people who like walking and running only, running and cycling only, and walking and cycling only.
=8+6+4=18= 8 + 6 + 4 = 18

3. Final Answer

(a) Venn Diagram: (See explanation above for the values in each region.)
(b) Total number of people in the locality: 39
(c) Number of people who like walking only: 10
(d) Number of people who like two different activities only: 18

Related problems in "Probability and Statistics"

We need to prove the Law of Total Probability, which states that if events $B_1, B_2, ..., B_n$ form...

ProbabilityLaw of Total ProbabilityConditional ProbabilitySet Theory
2025/4/9

The image provides a formula for calculating the probability of an event $A$, denoted as $P(A)$, usi...

ProbabilityLaw of Total ProbabilityConditional ProbabilityEvents
2025/4/9

The image describes the Law of Total Probability. Given a sample space $S$ of an experiment $E$, an...

ProbabilityLaw of Total ProbabilityConditional ProbabilityEventsSample SpacePartition
2025/4/9

We are given two sets of numbers. We need to find the five-number summary (minimum, first quartile, ...

Descriptive StatisticsFive-Number SummaryQuartilesMedianData Analysis
2025/4/8

The problem is to create a box plot for the dataset {9, 15, 19, 5, 20, 15}. We need to order the dat...

Box PlotData AnalysisQuartilesMedianDescriptive Statistics
2025/4/8

Jerry selects 20 boxes of crayons and finds that 2 of them have at least one broken crayon. If ther...

ProportionStatistical InferenceExpected Value
2025/4/8

The problem asks us to determine whether two given surveys are random or biased. - Problem 1: David ...

SamplingBiasRandomnessSurveys
2025/4/8

We are asked to calculate the probability of each of the six given events.

ProbabilityCard ProbabilitiesDice ProbabilitiesCoin Flip ProbabilitiesIndependent EventsComplementary Probability
2025/4/8

The problem provides a discrete probability distribution and asks to calculate the expected value $E...

Probability DistributionExpected ValueModeVarianceStandard DeviationDiscrete Probability
2025/4/8

The problem asks to construct the cumulative distribution function (CDF), $F(x)$, for a discrete ran...

Cumulative Distribution FunctionCDFDiscrete Random VariableProbability DistributionStep Function
2025/4/8