A special fair cubic die is rolled which has one blue side, two red sides and three yellow sides. The result is the colour of the top side after the die is rolled. If the die is rolled 8 times, then the probability of blue colour coming at least twice is

Options

A. `(13xx5^(7))/(6^(8))`
B. `(6^(8)-13xx5^(7))/(6^(8))`
C. `(8xx5^(7))/(6^(8))`
D. `1-(8xx5^(7))/(6^(8))`

Answer

To solve the problem of finding the probability of getting the blue color at least twice when rolling a special die 8 times, we can follow these steps:

### Step 1: Define the Probabilities
The die has the following sides:
- 1 blue side
- 2 red sides
- 3 yellow sides

The total number of sides = 1 + 2 + 3 = 6.

The probabilities of each color appearing when the die is rolled are:
- Probability of blue (P) = Number of blue sides / Total sides = 1/6
- Probability of not getting blue (Q) = 1 - P = 5/6

### Step 2: Define the Random Variable
Let X be the random variable representing the number of times blue appears when the die is rolled 8 times. We want to find the probability that X is at least 2, i.e., P(X ≥ 2).

### Step 3: Use the Complement Rule
To find P(X ≥ 2), we can use the complement rule:
\[ P(X ≥ 2) = 1 - P(X < 2) \]
This means we need to calculate \( P(X < 2) \), which includes the cases where X = 0 and X = 1.

### Step 4: Calculate P(X = 0) and P(X = 1)
Using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} P^k Q^{n-k} \]
where n is the number of trials (rolls), k is the number of successes (blue appearances), P is the probability of success, and Q is the probability of failure.

- For \( P(X = 0) \):
\[ P(X = 0) = \binom{8}{0} \left(\frac{1}{6}\right)^0 \left(\frac{5}{6}\right)^8 = 1 \cdot 1 \cdot \left(\frac{5}{6}\right)^8 = \left(\frac{5}{6}\right)^8 \]

- For \( P(X = 1) \):
\[ P(X = 1) = \binom{8}{1} \left(\frac{1}{6}\right)^1 \left(\frac{5}{6}\right)^7 = 8 \cdot \left(\frac{1}{6}\right) \cdot \left(\frac{5}{6}\right)^7 \]

### Step 5: Combine the Probabilities
Now we can combine these probabilities to find \( P(X < 2) \):
\[ P(X < 2) = P(X = 0) + P(X = 1) \]
\[ P(X < 2) = \left(\frac{5}{6}\right)^8 + 8 \cdot \left(\frac{1}{6}\right) \cdot \left(\frac{5}{6}\right)^7 \]

### Step 6: Calculate P(X ≥ 2)
Now we can find \( P(X ≥ 2) \):
\[ P(X ≥ 2) = 1 - P(X < 2) \]
\[ P(X ≥ 2) = 1 - \left(\frac{5}{6}\right)^8 - 8 \cdot \left(\frac{1}{6}\right) \cdot \left(\frac{5}{6}\right)^7 \]

### Step 7: Final Calculation
We can simplify the expression:
\[ P(X ≥ 2) = 1 - \left(\frac{5}{6}\right)^8 - \frac{8}{6} \cdot \left(\frac{5}{6}\right)^7 \]
\[ = 1 - \left(\frac{5}{6}\right)^7 \left(\frac{5}{6} + \frac{8}{6}\right) \]
\[ = 1 - \left(\frac{5}{6}\right)^7 \cdot \frac{13}{6} \]

Thus, the final probability is:
\[ P(X ≥ 2) = 1 - \frac{13 \cdot 5^7}{6^8} \]

### Final Answer
The probability of getting blue color at least twice when rolling the die 8 times is:
\[ P(X ≥ 2) = \frac{6^8 - 13 \cdot 5^7}{6^8} \]