$A$ man is known to speak the truth $2$ out of $3$ times. If he throws a die and reports that it is $6$,then the probability that it is actually $5$ is:

  • A
    $\frac{3}{8}$
  • B
    $\frac{1}{7}$
  • C
    $\frac{2}{7}$
  • D
    $\frac{4}{5}$

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Suppose we have four boxes $A, B, C$ and $D$ containing coloured marbles as given below:
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$B$ $6$ $2$ $2$
$C$ $8$ $1$ $1$
$D$ $0$ $6$ $4$

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