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What is the probability that, in six throws of a die, there will be exactly one each of “1”, “2”, “3”, “4”, “5” and “6”?

  1. $0.00187220$
  2. $0.01432110$
  3. $0.01176210$
  4. $0.01543210$
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The favorable event $E$ is $\{1, 2, 3, 4, 5, 6\}$ in any order. Also, it is evident that the event of getting one of $\{1, 2, 3, 4, 5, 6\}$ is independent.

On the first roll of the dice, any one of $\{1, 2, 3, 4, 5, 6\}$ can come up with a probability of $1$.

On the second roll, the favorable event is to get any number other than the one got in the first roll. This has a probability of $\frac{5}{6}$.

On the third roll, the favorable event is to get a number other than the one got in the first two rolls. This has a probability of $\frac{4}{6}$.

This continues $\cdots$

The required probability $P(E) = 1 \times \frac{5}{6} \times \frac{4}{6} \times \frac{3}{6} \times \frac{2}{6} \times \frac{1}{6} = 0.0154$
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4) 0.015 = $\frac{6!}{6^{6}}$ ??

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Your answer is correct, but any answer should be accompanied with an explanation. Don't you think so?
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