What is the average number of children Amy and Dan must​ have? (Integer or decimal. Do not round.)

Question

Amy and Dan want to have both a male and a female child. If they decide to continue to have children until they have one of each​ gender, what would the average number of children​ be? Assume that boys and girls are equally likely. Let the digits 0 to 4 represent a male child and the digits 5 to 9 represent a female child. Twenty trials were randomly generated. Use each of the numbers as a separate trial.

249

39

236

3437

105

45

690

6791

37

698884

7853

80

6753

29

72

51

62

64

24327

885791

Answer 1

2 or 3 digits most likely

as 2nd try has probability 1/2

3rd try probability 1/4. That's 3/4

one try has zero chance, since you can't get 1 of each on one try

4, 5, 6 or more digits have 1-3/4 = 1/4 chance

average for 20 tries?

13 have 2 or 3 digits

7 have 4 or more digits

7/20 = 35% chance of 4 or more digiits, 4th or more tries

13/20 = 65% chance of 2 or 3 digits, 2nd or 3rd try

experimental odds, not theoretical odds

Answer 2

The sum of lengths in the 20 trials is 63, which means a mean of 3.15

With this problem, we have 1 birth, either male or female, then a geometric distribution with p = 1/2 for the other gender. Thus, the mean is 1 + 1/p = 1 + 1/(1/2) = 1 + 2 = 3. The experimental result is reasonably close to the theoretical result.