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I'm reading the textbook entitled Harvard Statistics by Joseph K. Blitskein. In conditional probability context, I'm a little bit confused of calculating the intersection of event A and B $P(A\cap B)$. The problem is given below;

Example 2.2.5 (Elder is a girl vs. at least one girl). A family has two children, and it is known that at least one is a girl. What is the probability that both are girls, given this information? What if it is known that the elder child is a girl?

For the solution (based on textbook):

First, let the following events equal to A and B (just to keep it short):

A = event of having both girls, B = event of at least one girl

$$P(A|B)=\frac{P(A∩B)}{P(B)}=\frac{1/4}{3/4}=1/3$$

Intuitively, the answer is correct but I was wondering on why $P(A∩B)=1/4$. Since $P(A∩B)=P(A)P(B)$, which in this case $P(A)=1/4$ and $P(B)=3/4$. Just to give you a reference, the possible combinations of having two children are ${\{GG, BG, GB, BB}\}$.

Anyone can shed me a light regarding with my problem?

StubbornAtom
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King
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    Probabilty both girls and probability at least one girl are not independent. It is impossible to have two girls if there is not at least one girl and if you have two girls it is impossible not to have at least one girl. Common sense: $A\subset B$ and $A\cap B = A$ and $P(A\cap B) = P(A)$ – fleablood Nov 22 '19 at 02:16
  • Thing to realize is when $A \subset B$ then $A\cap B = A$ and $P(A\cap B) = P(A)$. – fleablood Nov 22 '19 at 02:19
  • related https://math.stackexchange.com/questions/1266043/boy-and-girl-paradox-is-driving-me-crazy – Henry Mar 04 '20 at 13:58

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$A$ and $B$ should be defined as events and not probabilities. $A$ is the event of having both girls and $B$ is the event of having at least one girl. By definition, you can see that they are not independent events. Hence $P(A,B) = P(A)P(B)$ does not hold. When we estimate $P(A,B)$, only one combination works $(GG)$ out of the four possible combinations. Hence the probability of both having both girls and having at least one girl is $P(A,B)=1/4$.

Henry
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Kaushik
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  • Oh I see! thanks for the correction. I misunderstood the definition of $P(A,B)$ that's the reason why I'm having a trouble on calculating the intersection of two events. This rule $P(A,B) = P(A)P(B)$ is only applicable for dependent events? – King Nov 22 '19 at 02:31
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    @King $P(A,B)=P(A)P(B)$ only for *in*dependent events. – David K Nov 22 '19 at 02:44
  • Thanks I got it now :) sorry for noob question. – King Nov 22 '19 at 02:50