A dice has 6 faces. 2 faces are colored in blue, 3 faces are colored in red and 1 is green. The dice is rolled 4 times. What is the probability of the dice to show: a) blue, red, red, green (in this order); b) The same colors but in any order?

a) Since rolling dices are independent events, we combine probability by multiplying them.

b) There are 3 colors, hence the probability of rolling individual colors is just its probability to the power of 4.

To find the probability of the dice showing specific sequences of colors, we'll first calculate the probability for each individual event and then combine them accordingly.

a) Probability of getting "blue, red, red, green" in this order:

The probability of rolling "blue" followed by "red" followed by another "red" and finally "green" is calculated by multiplying the probabilities of each event:

$P(\text{blue, red, red, green}) = P(\text{blue}) \times P(\text{red}) \times P(\text{red}) \times P(\text{green})$

Given that there are 6 faces in total and 2 of them are blue, 3 are red, and 1 is green:

$P(\text{blue}) = \frac{2}{6} = \frac{1}{3}$ $P(\text{red}) = \frac{3}{6} = \frac{1}{2}$ $P(\text{green}) = \frac{1}{6}$

Substituting these probabilities into the formula:

$P(\text{blue, red, red, green}) = \frac{1}{3} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{6}$

$P(\text{blue, red, red, green}) = \frac{1}{3} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{6} = \frac{1}{72}$

b) Probability of getting the same colors but in any order:

Here, we can use the concept of permutations because the order of colors matters. There are 4 rolls, with 2 blue, 3 red, and 1 green faces. We want to find the probability of getting the same colors but in any order.

Total number of possible outcomes = Total permutations of 4 rolls = $4! = 24$

Out of these, we need to find the permutations where we have 2 blue, 3 red, and 1 green face.

Number of permutations with 2 blue, 3 red, and 1 green face = $\frac{4!}{2! \times 2! \times 3!}$

$= \frac{24}{2 \times 2 \times 6}$ $= \frac{24}{24}$ $= 1$

Therefore, the probability of getting the same colors but in any order is 1 out of 24.

So, to summarize: a) The probability of rolling "blue, red, red, green" in that order is $\frac{1}{72}$. b) The probability of rolling the same colors but in any order is $\frac{1}{24}$.