Question

If 7 different balls are distributed among 4 different boxes, then the probability that the first box contains 3 balls is:

• A. \(\frac{35}{128}{(\frac{3}{4})}^{3}\)
• B. \(\frac{35}{64}{(\frac{3}{4})}^{4}\)
• C. \(\frac{7}{8}(\frac{3}{4})^{7}\)
• D. \(\frac{5}{16}(\frac{3}{4})^{5}\)

Hint:

For distributing $n$ different items into $k$ different boxes, the total number of ways is $k^n$. To choose $r$ items out of $n$, use the combination formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.

Answer 1

The Correct Option is B

Step 1: Determine the total number of ways to distribute the balls.
Each of the 7 balls can be placed into any of the 4 boxes.
Total number of ways = 4⁷.

Step 2: Determine the number of ways to select 3 balls for the first box.
We need to choose 3 balls out of 7 to be placed in the first box.
Number of ways to choose 3 balls = \(\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\).

Step 3: Determine the number of ways to distribute the remaining 4 balls.
The remaining 4 balls can be placed into any of the other 3 boxes.
Number of ways to distribute the remaining 4 balls = 3⁴ = 81.

Step 4: Calculate the number of favorable outcomes.
Favorable outcomes = \(\binom{7}{3} \times 3^4 = 35 \times 81\).

Step 5: Calculate the probability.
Probability = \(\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{35 \times 3^4}{4^7} = \frac{35 \times 81}{16384}\).
We can rewrite this as:
\(\frac{35 \times 81}{16384} = \frac{35 \times 3^4}{4^7} = \frac{35}{4^3} \times \frac{3^4}{4^4} = \frac{35}{64} \times (\frac{3}{4})^4\).
Therefore, the probability that the first box contains 3 balls is \(\frac{35}{64}(\frac{3}{4})^4\).