### Brain Teasers

# Family Problem

Probability
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.)

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.)

### Hint

Start by listing the possible families and determine the probability of each.e.g. Male-Male-Female-STOP. P(MMF) = (0.5)^3 = 0.125

A probability tree will help, for those who know how to construct one.

### Answer

2.75Possible families with 2 children:

M-F or F-M

The probability of each of these is (0.5)^2, = 0.25.

So 0.25 + 0.25 = 0.5 had 2 children.

Possible 3-child 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 4-child 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

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!]

really good teaser had to think about that one!! good job on being the first one

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...

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

.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

why the order id important in case of families with 2 children (M-F,F-M) are different, while this is not the case for families having 3 or more children

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