Genetics Problem

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shefv

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I see how they did the zero children with blue eyes and subtracted from 100%.

Isn't there a way to do this by calculating the probability for each child and then add that up?

Since the question stem states that "at least one of the next two children also have blue eyes", I interpreted that as being: child 1 has blue eyes (1/4), child 2 has blue eyes (1/4) or both have blue eyes (1/4*1/4) - all thee of these situations would fulfill the at least one child criteria.

So I did, 1/4 + 1/4 + (1/4*1/4) = 9/16 = 56%

What am I doing wrong? Can someone go over how to do this without the subtraction method as given in the solution? Is my interpretation of "at least one child" incorrect?

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If you do it your proposed way, you must account for all the possible states. Therefore,

1/4 * 3/4 + 1/4*3/4 + 1/4*1/4 = ~.44

This is like saying

P(1 blue 1 brown) + P (1 brown 1 blue) + P (2 blue)
 
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