Probability with dices involving colored sides

Question

We have three colored symmetrical dices each having 3 red sides 2 yellow sides and 1 green side. If they are rolled simultaneously what is the probability that.

a) Each dice show the same same color.

b) Each of the the dices show different colors.

My attempt at a)

Let $R$ be the event where all dices show red, then we have $P(R)=(\frac{1}{2})^3$.

Let $Y$ be the event that all dices show yellow, then we have $P(Y)=(\frac{1}{3})^3$.

Let $G$ be the event that all dices show green, then we have $P(G)=(\frac{1}{6})^3$

Since the events are independent we have the sought probability as $P(R\cup Y \cup G)=P(R)+P(Y)+P(G)=(\frac{1}{2})^3+(\frac{1}{3})^3+(\frac{1}{6})^3 \approx 16.6$%

My attempt at b)

Probability that a dice show red $P(D_R)=\frac{1}{2}$. Probability that a dice show yellow $P(D_Y)=\frac{1}{3}$. Probability that a dice show green $P(D_G)=\frac{1}{6}$.

Product rule gives $3!(P(D_R)P(D_Y)P(D_G))$ since these events can happen in $3!$ ways.

which is also $\approx 16.6$%.

Answer 1

You ask (in comments to the answer by aschepler) "Are my answers correct?".

Here's how I usually approach a probability problem where I am unsure of the correctness of my answer: I write a Monte Carlo simulation and run it many times to see if the results are consistent with my analytic result. Of course this requires basic skills in one programming language or another. Here is some code in R to simulate part a), simulating $10^5$ rolls of the three dice.

# ncolors: simulate roll of n special dice (3 red, 2 yellow, 1 green side)
# and return the number of distinct colors rolled
ncolors <- function(n) {
  # generate an array to hold the result of each roll
  colors <- rep(0, n)
  for (i in 1:n)
    # we encode red as 1, yellow as 2, green as 3
    colors[i] <- sample(c(1,1,1,2,2,3), 1)
  # we use the R "table" function to count the number of distinct colors
  return(length(table(colors)))
}
nreps <- 1e5
# set the random generator seed for reproducibility of results
set.seed(1234)
x <- replicate(nreps, ncolors(3) == 1)
print(sum(x))
prop.test(sum(x), nreps)

The result of runnning this code is that rolling the three dice resulted in a single color in $16614$ cases out of $10^5$, which yields a point estimate of $p = 0.16614$, with a 95% confidence interval (given by the R prop.test function) of $0.1638$ to $0.1685$. Since that confidence interval includes $1/6$, the Monte Carlo result is consistent with your analytic result.

Answer 2

Your answers and reasoning are good, except that in part (a) "independent" is the wrong word. The events are "mutually exclusive". And as mentioned in comments, it would be nice to show the answers in the $\frac{1}{6}$ form, not only the percent approximation.

Events are independent if the occurrence of one does not affect the probability of the other. Mathematically, $P(A \cap B) = P(A) P(B)$ and $P(A \cup B) = 1 - (1-P(A))(1-P(B))$.

Two events are mutually exclusive if it's impossible for both to happen. Mathematically, $P(A \cap B) = 0$ and $P(A \cup B) = P(A) + P(B)$.