In a given village there are only three sizes of families: families with 2 members, families with 4 members and families with 6 members. The proportion of families with 2, 4 and 6 members are roughly equal. A poll is conducted in this village wherein a person is chosen at random and asked about his/her family size. The average family size computed by sampling 1000 such persons from the village would be closest to

Let me break this down so it's crystal clear!

We have 3 types of families: 2-member, 4-member, and 6-member families with equal proportions. The key insight is that equal proportions of families ≠ equal chance of picking any person! We're picking people randomly, not families randomly.

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Let's say there are 300 families total (100 of each type for easy math):

• 100 families with 2 members = $100 \times 2 = 200$ people
• 100 families with 4 members = $100 \times 4 = 400$ people
• 100 families with 6 members = $100 \times 6 = 600$ people

Total people in village = $200 + 400 + 600 = 1200$ people

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Since we're picking people randomly:

• Probability of picking someone from 2-member family = $\dfrac{200}{1200} = \dfrac{1}{6}$
• Probability of picking someone from 4-member family = $\dfrac{400}{1200} = \dfrac{1}{3}$
• Probability of picking someone from 6-member family = $\dfrac{600}{1200} = \dfrac{1}{2}$

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When we ask each person about their family size:

• People from 2-member families will say "2"
• People from 4-member families will say "4"
• People from 6-member families will say "6"

Expected average = (Probability × Family Size) for each type:

$= \left(\dfrac{1}{6} \times 2\right) + \left(\dfrac{1}{3} \times 4\right) + \left(\dfrac{1}{2} \times 6\right)$

$= \dfrac{2}{6} + \dfrac{4}{3} + 3$

$= \dfrac{1}{3} + \dfrac{4}{3} + \dfrac{9}{3}$

$= \dfrac{14}{3} = 4.67$

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The average family size from sampling 1000 people would be closest to 4.67.