Key Concepts and Formulas

• Binomial Distribution: The probability of $k$ successes in $n$ independent trials, each with probability of success $p$, is given by $P(X=k) = {^nC_k} p^k (1-p)^{n-k}$, where ${^nC_k} = \frac{n!}{k!(n-k)!}$.
• Expected Value of a Binomial Distribution: If $X$ follows a binomial distribution with $n$ trials and probability of success $p$, then the expected value of $X$ is $E(X) = np$.
• Complement Rule: $P(A) = 1 - P(A')$, where $A'$ is the complement of event $A$.

Step-by-Step Solution

Step 1: Define the event of interest and calculate the probability of success for a single die.

Why: We need to determine the probability of a single die showing a 3 or a 5, as this will be the basis for calculating the probabilities for multiple dice. Let's define success as rolling a 3 or a 5 on a single die. The sample space for a single die is {1, 2, 3, 4, 5, 6}, and the favorable outcomes are {3, 5}. The probability of success for a single die, $p_d$, is: $p_d = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{6} = \frac{1}{3}$ The probability of failure (not rolling a 3 or a 5) is: $q_d = 1 - p_d = 1 - \frac{1}{3} = \frac{2}{3}$

Step 2: Define the event of interest for a single throw of four dice.

Why: We need to define what constitutes a "success" in the context of throwing four dice, which is "at least two dice show a 3 or a 5". Let $X$ be the number of dice (out of four) that show a 3 or a 5. Since each die is independent, $X$ follows a binomial distribution with parameters $n_1 = 4$ (number of dice) and $p_1 = p_d = \frac{1}{3}$ (probability of success for a single die). We want to find the probability that at least two dice show a 3 or a 5, which is $P(X \ge 2)$.

Step 3: Calculate the probability of the event of interest for a single throw of four dice, $P(X \ge 2)$.

Why: This step calculates the probability of success ($P(X \ge 2)$) for one trial (throwing four dice), which is crucial for calculating the expected number of successes over 27 trials. It's easier to calculate the complement: $P(X \ge 2) = 1 - P(X < 2) = 1 - [P(X=0) + P(X=1)]$ Using the binomial probability formula $P(X=k) = {^nC_k} p^k (1-p)^{n-k}$, we calculate $P(X=0)$ and $P(X=1)$: $P(X=0) = {^4C_0} \left(\frac{1}{3}\right)^0 \left(\frac{2}{3}\right)^4 = 1 \cdot 1 \cdot \frac{16}{81} = \frac{16}{81}$ $P(X=1) = {^4C_1} \left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^3 = 4 \cdot \frac{1}{3} \cdot \frac{8}{27} = \frac{32}{81}$ Therefore: $P(X \ge 2) = 1 - \left(\frac{16}{81} + \frac{32}{81}\right) = 1 - \frac{48}{81} = 1 - \frac{16}{27} = \frac{27-16}{27} = \frac{11}{27}$

Step 4: Calculate the expected number of times the event occurs in 27 throws.

Why: We now have the probability of success for a single experiment (throwing four dice). We use this probability to calculate the expected number of successes over 27 independent repetitions of the experiment. We are throwing the four dice 27 times independently. Let $Y$ be the number of times that at least two dice show a 3 or a 5. $Y$ follows a binomial distribution with parameters $n_2 = 27$ (number of trials) and $p_2 = P(X \ge 2) = \frac{11}{27}$ (probability of success in a single trial). The expected number of times this event occurs is: $E(Y) = n_2 \cdot p_2 = 27 \cdot \frac{11}{27} = 11$

Common Mistakes & Tips

• Confusing the two binomial distributions: There is one binomial distribution for the number of dice showing a 3 or a 5 in a single throw of four dice, and another for the number of times the event occurs in 27 throws.
• Using the complement rule effectively: Calculating $P(X \ge 2)$ directly would require calculating $P(X=2)$, $P(X=3)$, and $P(X=4)$, which is more work than calculating $P(X=0)$ and $P(X=1)$ and using the complement rule.
• Simplifying fractions: Always simplify fractions to make calculations easier.

Summary

We calculated the probability of at least two dice showing a 3 or a 5 in a single throw of four dice. Then, using this probability, we calculated the expected number of times this event occurs when the four dice are thrown 27 times independently. The expected number of times is 11.

Final Answer

The final answer is $\boxed{11}$.