$12$ balls are distributed among $3$ boxes. The probability that the first box will contain exactly $3$ balls is:

  • A
    $\frac{{}^{12}C_3 \times 2^9}{3^{12}}$
  • B
    $\frac{{}^{12}C_3 \times 2^9}{3^{10}}$
  • C
    $\frac{{}^{12}C_3}{3^{12}}$
  • D
    $\frac{{}^{12}C_3}{3^{10}}$

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If $A$ and $B$ are independent events and $P(A) = p, P(B) = 2p$,and $P(\text{exactly one from } A \text{ and } B) = \frac{5}{9}$,then find the value of $p$.

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