Q75 / 160 · AP EAMCET 2024 (19 May Shift 2) · Mathematics

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

  • A
    $\frac{35}{128}\left(\frac{3}{4}\right)^3$
  • Correct answer
    B
    $\frac{35}{64}\left(\frac{3}{4}\right)^4$
  • C
    $\frac{7}{8}\left(\frac{3}{4}\right)^7$
  • D
    $\frac{5}{16}\left(\frac{3}{4}\right)^5$

Solution

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 $=\frac{{ }^7 C_3 \times 3^4}{4^7}=\frac{35}{64}\left(\frac{3}{4}\right)^4$.

Hence, the correct answer is option: 'B'.

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