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Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

 

Puzzle ID:#6367
Fun:*** (2.2)
Difficulty:*** (2.78)
Category:Probability
Submitted By:cathalmccabe***
Corrected By:BobbyKat

 

 

 



The chances of a mother giving birth to a son compared with a daughter are 55%-45% (This is true in the real world due to a higher male infant mortality rate although the percent might be closer together depending on the type of country. 1st world, 3rd world - more males are born than females)

In a family of four children, what are the odds a mother will have 2 boys and a girl?


Answer

I've listed all the possibilities b = boy g = girl
The chance of a child being male/female is independent, but the chance of having a certain order is just the probability multiplied by each other.
Because one child doesn't matter; the desired scenarios are any with either 3 boys 1 girl or 2 girls 2 boys.

I've marked with an X the desired outcomes, and at the end I have the probability of each occurrence.
Simply add the probabilities together for each desired case!

(as an eg of the calculation, bbgg is 0.55*0.55*0.45*0.45 which is the same answer as ggbb/ bgbg)

b bbb x 0.09150625
b bbg x 0.07486875
b bgb x 0.07486875
b bgg x 0.06125625
b gbb x 0.07486875
b gbg x 0.06125625
b ggb x 0.06125625
b ggg x 0.05011875
g bbb x 0.07486875
g bbg x 0.06125625
g bgb x 0.06125625
g bgg x 0.05011875
g gbb x 0.06125625
g gbg x 0.05011875
g ggb x 0.05011875
g ggg x 0.04100625



Answer is 0.6670125
or approx 2/3


(As a check all the probabilities should sum 1)

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