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Step-by-Step Solution
Step 1: Identify the Total Number of Ways to Distribute the Balls
There are 9 distinct balls and 4 different boxes (B1, B2, B3, and B4).
Each ball can go into any one of the 4 boxes independently. Thus, the total number of ways to distribute
all 9 balls into 4 boxes is
$4^9$.
Step 2: Count the Favorable Ways Where Box B3 has Exactly 3 Balls
1. Choose which 3 balls go into box B3. There are $\binom{9}{3}$ ways to choose 3 balls out of 9.
2. The remaining 6 balls must be distributed among the other 3 boxes (B1, B2, and B4).
Each of these 6 balls can go into any of these 3 boxes, giving $3^6$ ways.
Therefore, the total number of favorable ways is
$\binom{9}{3} \times 3^6$.
Step 3: Compute the Probability
The probability that B3 has exactly 3 balls is the ratio of favorable ways to the total ways:
$$
\frac{ \binom{9}{3} \, 3^6 }{4^9}.
$$
Step 4: Express the Probability in the Form $k \left(\frac{3}{4}\right)^9$
We factor out $\left(\frac{3}{4}\right)^9$:
$$
\frac{ \binom{9}{3} \, 3^6 }{4^9}
\;=\; \binom{9}{3} \;\frac{3^6}{4^9}
\;=\; \binom{9}{3} \;\frac{1}{27}\;\left(\frac{3}{4}\right)^9
\;=\;\frac{\binom{9}{3}}{27}\;\left(\frac{3}{4}\right)^9.
$$
Since $\binom{9}{3} = 84$, we get
$$
\frac{84}{27}\left(\frac{3}{4}\right)^9 \;=\;\frac{28}{9}\left(\frac{3}{4}\right)^9.
$$
Hence, $k = \frac{28}{9}$.
Step 5: Verify Which Set Contains $k = \frac{28}{9}$
Numerically, $\frac{28}{9} = 3.111\ldots$, which must satisfy one of the given sets.
Note that $3.111\ldots$ is in the interval $(2,4)$, i.e.,
$|\,k - 3\,| < 1$.
Therefore, the correct set is $\{ x \in \mathbb{R} \colon |x - 3| < 1 \}$.
