Family Problem
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
The relatively unknown Mbwatzeze Tribe had a rather strange law restricting the size of families. Each married couple was expected to continue having children until they had EITHER one child of each sex OR a total of four children.
What was the average (mean) number of children born to each couple? (Assume that they kept to the law and that each couple had the maximum permitted by the law.)
HintStart by listing the possible families and determine the probability of each.
e.g. MaleMaleFemaleSTOP. P(MMF) = (0.5)^3 = 0.125
A probability tree will help, for those who know how to construct one.
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Answer
2.75
Possible families with 2 children:
MF or FM
The probability of each of these is (0.5)^2, = 0.25.
So 0.25 + 0.25 = 0.5 had 2 children.
Possible 3child families:
MMF or FFM
The probability of each of these is (0.5)^3, = 0.125.
So 0.125 + 0.125 = 0.25 had 3 children.
Possible 4child families:
MMMM, MMMF, FFFF or FFFM
The probability of each of these is (0.5)^4, = 0.0625.
So 0.0625 x 4 = 0.25 had 4 children.
Half the couples had 2 children, a quarter had 3 and a quarter had 4. So the average (mean) size is
0.5 x 2 + 0.25 x 3 + 0.25 x 4 = 2.75
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Comments
dalfamnest
Oct 23, 2009
 Hi everyone! Hope you enjoy this little teaser. I thought it was a fun idea. How about some comments, please  it's my first Probability Teaser and I'd value some DEEF, + or . [See today's TOTD!] 
valencia
Oct 23, 2009
 really good teaser had to think about that one!! good job on being the first one 
precious1026
Nov 15, 2009
 It seems to me the answer is difficult to achieve without certain information. Perhaps, it's the formation of the Question, which appears to assert there is more families than the one given in the problem. Well, I guess the means of 2.75 is a logical number of children in a family. I would say, and I usually say "keep them coming, but I would be lying if I did say "keep them coming. What is the probability of you making up more problems such as this? Actually, this is a good teaser for those who understand the mean... 
javaguru
Dec 30, 2009
 Simple and fairly easy. You can simplify it further by ignoring male/female permutations as:
.5 probability second child is different than first (2 kids);
.5 x .5 = .25 that second child is same as first and third child is different than first (3 kids);
.5 x .5 = .25 that second and third children are same as first (4 kids).
.5 x 2 + .25 x 3 + .25 x 4 = 2.75

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