Group A contains $17$ girls and $3$ boys. Group B contains $5$ girls $10$ boys. Suppose we randomly take $2$ students from group B to group A. Now we are choose one student: what is the probability that it is a boy and what is the probability that it is a girl? My attempts: P(boy)= $5/22+4/22+3/22=6/11$ .....why it isn't this right?
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Hint: Let $P(B)$ be the probability a boy is chosen from group A. The possible groupings of transfer students are: boy girl ($BG$), girl girl ($GG$) and boy boy ($BB$). Then
$$P(B) = P(B|BG)P(BG) + P(B|GG)P(GG) + P(B|BB)P(BB)$$
In your answer, you have not multiplied each case by the probability that each case occurs.
answered Nov 16, 2014 at 5:17
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$\begingroup$ @ AWertheim in the case of P(BG) is it suppose to be [(5/15)*(10/14)+(10/15)/(5/14)]? $\endgroup$ Nov 16, 2014 at 8:10
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$\begingroup$ @user188750: yep, it's indeed $[(5/15)*(10/14)+(10/15)*(5/14)]$, or $P(B|G)P(G)+P(G|B)P(B)$. Another way to think of it (which may be a bit simpler) is this: there are $5$ girls to pick from and $10$ boys. You want to choose one of each - the number of ways to do that is ${5 \choose 1} \cdot {10\choose 1} =50$. The total ways to choose two people is ${15 \choose 2}=105$, yielding $P(BG) = 50/105 \approx .476$. You can verify that these yield the same answer. $\endgroup$ Nov 16, 2014 at 18:16