Three different dice are rolled simultaneously, three times simultaneously. The find the probability that all of them show different numbers exactly two times.
Will total cases will be $(6^3)^3$? For favorable cases I think we can select two cases when we get different numbers on all three dice, then we will have $6 \times5 \times 4 $ favorable cases for these two cases, then in last cases at least two numbers must be same. Is my approach correct? Could someone help me with this?
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For any one roll, examining the dice one by one, the 1st will have some number 6/6 or 100% of the time. The second will differ from the first 5/6 of the time and the third will differ from the first two 4/6 of the time. So the probability of all being different is 6/6 * 5/6 * 4/6, or about 55.56%. The probability of two or three having the same number is about 44.44%.
For three rolls with one having 2 or 3 dice with the same number, and the other 2 rolls having all numbers different, the roll with repeated numbers might be the first second or third. So the probability of that outcome is 3 * .5556 * .5556 * .4444, or 41.15%
Let's assume the dice are distinguishable. Say they are blue, green, and red.
For a single roll of three dice, there are $6^3$ possible outcomes. There are $6 \cdot 5 \cdot 4$ ways to obtain three different outcomes (for each of the six outcomes of the blue die, there are five outcomes of the green die that are different from the outcome of the blue die, and four outcomes of the red die that are different from both the outcomes on the blue and green dice). Hence, the probability of obtaining three different outcomes when three dice are rolled simultaneously is $$p = \frac{6 \cdot 5 \cdot 4}{6^3}$$ What we want is the probability that in exactly three trials of rolling three dice simultaneously, exactly two of those trials produce three different outcomes on the dice. We can use the binomial distribution, which states that the probability of exactly $k$ successes in $n$ trials, each of which has probability $p$ of occurring is $$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$ where $\binom{n}{k}$ is the number of ways exactly $k$ successes could occur in $n$ trials, $p^k$ is the probability of $k$ successes, and $(1 - p)^k$ is the probability of $n - k$ failures. In your problem, $n = 3$, $k = 2$, and $p$ was worked out above.