Four different dice are thrown independently 27 times, then find the expectation of number of times if at leat two of them shows either 5 or 3.

Answer

To solve the problem of finding the expectation of the number of times at least two out of four different dice show either a 5 or a 3 when thrown independently 27 times, we can follow these steps:

### Step 1: Determine the Probability of a Die Showing 5 or 3
Each die has 6 faces, and the favorable outcomes for our event are showing either a 5 or a 3. Thus, the probability \( P \) of a die showing either a 5 or a 3 is:

\[ P(\text{5 or 3}) = \frac{2}{6} = \frac{1}{3} \]

### Step 2: Determine the Probability of a Die Not Showing 5 or 3
The probability \( Q \) of a die not showing either a 5 or a 3 is:

\[ Q = 1 - P = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 3: Define the Random Variable
Let \( X \) be the random variable representing the number of dice (out of 4) that show either a 5 or a 3. We want to find \( P(X \geq 2) \).

### Step 4: Use the Complement Rule
To find \( P(X \geq 2) \), we can use the complement rule:

\[ P(X \geq 2) = 1 - P(X < 2) = 1 - (P(X = 0) + P(X = 1)) \]

### Step 5: Calculate \( P(X = 0) \)
Using the binomial probability formula:

\[ P(X = 0) = \binom{4}{0} P^0 Q^4 = 1 \cdot \left(\frac{1}{3}\right)^0 \cdot \left(\frac{2}{3}\right)^4 = \left(\frac{2}{3}\right)^4 = \frac{16}{81} \]

### Step 6: Calculate \( P(X = 1) \)
Similarly, we calculate \( P(X = 1) \):

\[ P(X = 1) = \binom{4}{1} P^1 Q^3 = 4 \cdot \left(\frac{1}{3}\right)^1 \cdot \left(\frac{2}{3}\right)^3 = 4 \cdot \frac{1}{3} \cdot \frac{8}{27} = \frac{32}{81} \]

### Step 7: Combine the Probabilities
Now we can combine these probabilities to find \( P(X < 2) \):

\[ P(X < 2) = P(X = 0) + P(X = 1) = \frac{16}{81} + \frac{32}{81} = \frac{48}{81} \]

### Step 8: Find \( P(X \geq 2) \)
Now we can find \( P(X \geq 2) \):

\[ P(X \geq 2) = 1 - P(X < 2) = 1 - \frac{48}{81} = \frac{33}{81} = \frac{11}{27} \]

### Step 9: Calculate the Expectation
Since the dice are thrown 27 times, the expected number of times at least 2 dice show either a 5 or a 3 is:

\[ E = 27 \cdot P(X \geq 2) = 27 \cdot \frac{11}{27} = 11 \]

### Final Answer
Thus, the expectation of the number of times at least two of them show either 5 or 3 is:

\[ \boxed{11} \]