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AP EAMCET · Maths · Probability

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

  1. A \(\frac{35}{128}\left(\frac{3}{4}\right)^3\)
  2. B \(\frac{35}{64}\left(\frac{3}{4}\right)^4\)
  3. C \(\frac{7}{8}\left(\frac{3}{4}\right)^7\)
  4. D \(\frac{5}{16}\left(\frac{3}{4}\right)^5\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{35}{64}\left(\frac{3}{4}\right)^4\)

Step-by-step Solution

Detailed explanation

Since, total number of ways of distributing of 7 balls in 4 box \(=4^7\) Total number of ways such that first box contain 3 balls \(={ }^7 C_3 \times 3^4\) So, required probability \(=\frac{{ }^7 C_3 \times 3^4}{4^7}=\frac{35}{64}\left(\frac{3}{4}\right)^4\).