Given a pool of 12 objects divided into 5 groups (see table), what is the probability that, when 5 objects are randomly chosen from the pool, exactly 3 of them belong to group A or exactly 3 of them belong to group B?
Group | Objects
A | 3
B | 4
C | 2
D | 1
E | 2
My current solution involves finding the independent probabilities of choosing 3 from group A and from group B and adding them together:
$$ \begin{align} P(3 \subset A) &= \frac3{12} \cdot \frac2{11} \cdot \frac1{10} \cdot \frac99 \cdot \frac88 \\ &= \frac14 \cdot \frac2{11} \cdot \frac1{10} \\ &= \frac1{220} \\ \\ P(3 \subset B) &= \frac4{12} \cdot \frac3{11} \cdot \frac2{10} \cdot \frac89 \cdot \frac78 \\ &= \frac13 \cdot \frac3{11} \cdot \frac15 \cdot \frac89 \cdot \frac78 \\ &= \frac7{495} \\ \\ P(3 \subset A) + P(3 \subset B) &= \frac1{220} + \frac7{495} \\ &= \frac{37}{1980} \end{align} $$
Though my textbook says the correct answer is $ \frac{13}{66} $, and provides no process for the solution. Is my solution correct, and if not, how is the number given by my textbook calculated?
There are $\binom{12}{5}$ ways to select five of the twelve objects in the pool.
Let $A$ be the event that exactly three objects are selected from group A when five objects are selected from the pool of twelve objects. Let $B$ be the event that exactly three objects are selected from group B when five objects are selected from the pool of twelve objects.
The probability that exactly three objects are selected from group A or exactly three objects are selected from group B when five objects are selected from the pool of twelve objects is $$\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B)$$ Observe that $\Pr(A \cap B) = 0$ since it is not possible for a selection of five objects to contain exactly three objects from group A and three objects from group B since $3 + 3 = 6 > 5$.
For event $A$ to occur, all three objects in group A must be selected together with $2$ of the other $12 - 3 = 9$ objects in the pool, which can occur in $$\binom{3}{3}\binom{9}{2}$$ ways. Hence, $$\Pr(A) = \frac{\dbinom{3}{3}\dbinom{9}{2}}{\dbinom{12}{5}}$$
For event $B$ to occur, three of the four objects in group B must be selected together with $2$ of the other $12 - 4 = 8$ objects in the pool, which can occur in $$\binom{4}{3}\binom{8}{2}$$ ways. Hence, $$\Pr(B) = \frac{\dbinom{4}{3}\dbinom{8}{2}}{\dbinom{12}{5}}$$
Since events $A$ and $B$ are mutually exclusive, $$\Pr(A \cup B) = \Pr(A) + \Pr(B) = \frac{\dbinom{3}{3}\dbinom{9}{2} + \dbinom{4}{3}\dbinom{8}{2}}{\dbinom{12}{5}} = \frac{148}{792} = \frac{37}{198}$$ which means the answer in the textbook is incorrect.