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Consider families with three children. Assume that all the 8 possibilities corresponding to the gender of the children are equally likely. Let A denote the event that “There is at most one girl in the family”. Let B denote the event “The family has children of both genders”. Then, the events A and B are independent?

My Solution: Now if we consider event A occurs then : 1. number of girl child will be 0 2. number of girl child will be 1

Now if number of girl child is 0 then P(B) becomes 0 and if number of girl child is 1 then P(B)=1.

Similarly,if number of girl child =2 then P(B)=1 and number of girl child is 3 then P(B)=0

So how these events are independent.

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The two events are independent.

If two events $A$ and $B$ are independent, then by definition, $P(A\cap B)=P(A)\cdot P(B)$.

The probability of $A$ (the family having at most one girl) is $1/2$, since there is one way to have no girls and three ways to have one girl in the family, out of the eight possible cases.

The probability of $B$ (the family has both genders) is $3/4$, since there are only two ways to have only one gender in the family, out of the eight cases.

The probability of $A\cap B$ (the family has at most one girl and has both genders) is $3/8$, since the three cases with one girl ensures the family has both genders.

Checking $P(A\cap B)=P(A)\cdot P(B)$, we have $3/8\overset{?}{=}(1/2)(3/4)$, which is true. Hence, by the definition of independent events, $A$ and $B$ are independent.

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Since you have already received two detailed correct answers I will try to address what I think might be confusing about the conditions in this question.

You are given two events, one of them $B$ is gender-indifferent (it describes an event which says the same thing about both genders), whereas the other one, $A$, breaks the gender symmetry, and it seems intuitively impossible for two such events to be independent.

However, in the specific conditions of the question it is worth noting that the complement of $A$ is the event in which the family has at least two daughters, which is the same as saying it has at most one son, and is therefore the same as $A$ after switching the genders. So whether $A$ occurs or not is equivalent to choosing between two gender-symmetrical alternatives, and hence cannot change the probability of another, gender-symmetrical event, such as $B$.

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Independent probability events are defined by:

$$P(A\cap B)=P(A)P(B)$$

If this relation is satisfied, then the events are independent, if not, they are dependent.

So we can see that $A\cap B=\{GBB, BGB, BBG\}$ which gives $P(A\cap B)=\dfrac38$ and $P(A)P(B)=\dfrac12.\dfrac68=\dfrac38$, and so event $A$ and event $B$ are independent.

If we create an event $C$ - one of the first two children born was a girl, then we can see that $P(A\cap C)=\dfrac28$ whereas $P(A)P(C)=\dfrac38$, so these events are dependent.

$\begin{array} {c|c|c|c} children&A&B&C\\ \hline GGG&&&\checkmark\\ GGB&&\checkmark&\checkmark\\ GBG&&\checkmark&\checkmark\\ GBB&\checkmark&\checkmark&\checkmark\\ BGG&&\checkmark&\checkmark\\ BGB&\checkmark&\checkmark&\checkmark\\ BBG&\checkmark&\checkmark&\\ BBB&\checkmark&& \end{array} $

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  • $\begingroup$ Looking at your chart, $P(A)$ appears to be $1/2$, not $1/4$. $\endgroup$ – Tim Thayer Sep 24 '15 at 5:59
  • $\begingroup$ @TimThayer; just spotted that myself - fixing... $\endgroup$ – JMP Sep 24 '15 at 6:02

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