I don't understand how to solve problems of conditional probability.Here's a problem which I fail to understand."Consider all families with two children and assume that each child is equally likely to be a boy or a girl.If such a family is chosen at random and found to have a boy,then what is the probability that it has another boy."Please help me to solve this problem.
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1Try this for a start. – David Oct 19 '15 at 04:38
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Let $A$ be the event that the family have a boy and $B$ be the event that the family has another boy.
Then $P(B|A)=P(AB)/P(A)=(1/4)/(3/4)=(1/3)$
Esther Jacob
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How P(A)=3/4?It should be 2/3.Because the combination (boy,girl) and (girl,boy) is symmetric.Is it not right? – Arnab Chattopadhyay Oct 19 '15 at 05:00
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@ArnabChattopadhyay Since the sample space is $S={BB,BG,GB,GG}$, there are $3$ of them in event $A$ – Esther Jacob Oct 19 '15 at 05:04
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When we say that the probability of an event $A$ is conditioned on an event $B$, we can think of this intuitively as, what's the chance that $A$ happens knowing $B$ is something that already happened?
You might hopefully be familiar with the expression with $P(A|B)$ being the probability of $A$ given $B$:
$P(A|B)=\frac{P(A \cap B)}{P(B)}$.
As you can see, we're considering the event of $A$ and $B$ happening, divided by $P(B)$. You can think of this as $P(B)$ becoming our 'sample space' as it is the event of interest regarding $A$'s probability, what are the odds that $A$ occurs if $B$ occurs? Why this sample space intuition comes around is simply that:
$P(A|B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{|A \cap B|}{|S|}}{\frac{|B|}{|S|}}=\frac{|A \cap B|}{|S|}\times \frac{|S|}{|B|}=\frac{|A \cap B|}{|B|}$, where $S$ is our sample space.
Let's model the problem, as it is good to do so. What is our sample space?
$S=${$BB,BG,GB,GG$}, where $G$ denotes a girl, and $B$ denotes a boy. You don't necessarily have to write out the sample space but this is to get a better understanding of what's happening. Be aware that the gender of each child is independent of one another, so you will see as to why we ought to consider $BG$ and $GB$ to be in our sample space.
Then let $B_1$ be the event that one of children of the family is a boy. This is important. And let $B_0$ denote the event that the other child is a boy.
We seek $P(B_0|B_1)$ = $P(B_0 \cap B_1)/P(B_1)$. This is the probability that the other child is a boy in the family given that one of them is a boy.
Can you piece together the end result?
Areyouheaty
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