Probability Paradox Solution


Which lady has a higher probability that both 
children are girls? Or are the probabilities 
equal? 


Answer:  Lady A has a higher probability that both are girls.
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Lady A: "My oldest child (of two) is a girl."


If her oldest child is a girl, then we know 
she has one girl. So the probability of her 
second child being a girl is 1/2. It's like 
a conditional probability. Suppose X is the 
event of her first child and Y is the event of 
her second child. We want to find out 
P(Y=girl | X=girl). This means, what is the 
probability of the second child being a girl 
given the fact that her first child is a girl.
Well if you know a little probability theory, you 
know that

                 P[X=a and Y=b]
P[X=a | Y=b] =  ----------------
                     P[Y=b]

So this translates to:
(Here we use the fact that the probability of 
having two girls, independently, is 1/4)


                       P[Y=girl and X=girl]     1/4
P[Y=girl | X=girl] =  ---------------------- = ---- = 1/2
                             P[X=girl]          1/2




Lady B: "At least one of my two children is a girl."


This is a little bit trickier. First, you should know 
that if X is an event, then the probability of X NOT 
happening is 1-(probability of X happening). This is 
because probabilities always add up to 1. So first note 
that the probability of at least one child being a girl 
is the same as 1-(probability that no children are girls).
This is the same as 1-(probability that both are boys). 
But, logically, the probability that two kids are boys is 
1/4. So 1-1/4 = 3/4. So the probability of at least one 
of the two children being a girl is 3/4. But now we want 
to find the probability of two girls. So our "universal 
space" is 3/4. But we know that the probability of two girls
is normally the same as two boys, 1/4. However, we have to 
divide out or universal space. So,
               
                  1/4
P[two girls] =   ----- = 1/3
                  3/4


jmg 6/10/05