Final answer:
Using binomial probability, the expected number of families with at least one boy is 937, with no girl is 62, and with at most 2 girls is 687. However, these numbers do not match exactly with the options given, though option (d) is the closest.
Step-by-step explanation:
The question pertains to the probability of certain family compositions in a population of 1000 families with each having 4 children. We will use the concept of binomial probability to solve this.
(i) At least one boy: The only scenario where there isn't at least one boy is when all children are girls. The probability of having a girl is 1/2, so the probability of having 4 girls in a row is (1/2)^4 = 1/16. Therefore, the probability of having at least one boy is 1 - 1/16 = 15/16. Out of 1000 families, the expected number would be 1000 * (15/16) = 937.5, which we would round down to 937 since we can't have a fraction of a family.
(ii) No girl: This is the probability of having all boys which is again (1/2)^4 = 1/16. Thus, expected number of families with no girls is 1000 * (1/16) = 62.5, rounded down to 62 families.
(iii) At most 2 girls: This includes the probability of having 0, 1, or 2 girls. Using binomial probability, P(0 girls) + P(1 girl) + P(2 girls) which is (1/16) + (4/16) + (6/16) = 11/16. Therefore, the expected number of families with at most 2 girls is 1000 * (11/16) = 687.5, rounded to 687 families.