Four fair dice are thrown independently 27 times. Then the expected number of times, at least two dice show up a three or a five, is

We throw four fair dice in one go and we are interested in the event that “at least two dice show a three or a five”. Let us first analyse a single throw of the four dice and then extend the result to the 27 independent repetitions.

For any single die, a face can be $$1,2,3,4,5,6$$, all equally likely. The favourable faces for our event are $$3$$ or $$5$$, so for one die we have

$$P(\text{face is }3\text{ or }5)=\frac{2}{6}=\frac13.$$

Correspondingly,

$$P(\text{face is neither }3\text{ nor }5)=1-\frac13=\frac23.$$

Let the random variable $$X$$ denote “the number of dice (out of four) that show a 3 or a 5” in one single throw of the four dice. Because each die acts independently, $$X$$ follows a binomial distribution with parameters $$n=4$$ and $$p=\frac13$$. Symbolically,

$$X\sim\text{Binomial}(4,\tfrac13).$$

We need the probability that at least two dice among the four are favourable, i.e.

$$P(\,X\ge 2\,)=1-P(\,X=0\,)-P(\,X=1\,).$$

Using the binomial probability formula $$P(X=k)=\binom{n}{k}p^{\,k}(1-p)^{\,n-k},$$ we compute each term step-by-step:

First, $$P(X=0)=\binom{4}{0}\Bigl(\tfrac13\Bigr)^{0}\Bigl(\tfrac23\Bigr)^{4}=1\cdot1\cdot\Bigl(\tfrac23\Bigr)^{4}=\left(\frac23\right)^{4}=\frac{16}{81}.$$

Next, $$P(X=1)=\binom{4}{1}\Bigl(\tfrac13\Bigr)^{1}\Bigl(\tfrac23\Bigr)^{3}=4\cdot\frac13\cdot\left(\frac23\right)^{3}=4\cdot\frac13\cdot\frac{8}{27}=4\cdot\frac{8}{81}=\frac{32}{81}.$$

Now we combine these values:

$$P(X\ge 2)=1-\frac{16}{81}-\frac{32}{81}=1-\frac{48}{81}=\frac{33}{81}=\frac{11}{27}.$$

This probability represents the chance of a “success” (at least two favourable faces) in one four-dice throw. Let us denote this success probability by $$p_s=\frac{11}{27}.$$

We are told that the four dice are thrown independently a total of 27 times. Let the random variable $$Y$$ count “the number of throws (out of 27) in which at least two dice show a three or a five.” Each throw is independent and has success probability $$p_s$$, so

$$Y\sim\text{Binomial}(27, p_s).$$

For a binomial random variable, the expected value is given by the well-known formula

$$E(Y)=n\,p,$$

where $$n$$ is the number of trials and $$p$$ the success probability per trial. Substituting $$n=27$$ and $$p=p_s=\frac{11}{27},$$ we obtain

$$E(Y)=27\left(\frac{11}{27}\right)=11.$$

So, on average, we expect the desired event to occur eleven times in the 27 repetitions.

So, the answer is $$11$$.