For one roll of a certain number cube with six faces, numbered 1 through 6, the probability of rolling a two is `(1)/(6)`. If this number cube is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?

Options

A. `((1)/(6))^(4)`
B. `2((1)/(6))^(3)+((1)/(6))^(4)`
C. `3((1)/(6))^(3)((5)/(6))+((1)/(6))^(4)`
D. `4((1)/(6))^(3)((5)/(6))+((1)/(6))^(4)`

Answer

To solve the problem of finding the probability of rolling a two at least 3 times when a six-faced die is rolled 4 times, we can break it down into a few steps:

### Step 1: Define the probabilities
The probability of rolling a two (success) on a single roll of the die is:
\[ P(\text{two}) = \frac{1}{6} \]
The probability of not rolling a two (failure) is:
\[ P(\text{not two}) = 1 - P(\text{two}) = 1 - \frac{1}{6} = \frac{5}{6} \]

### Step 2: Identify the scenarios
We need to find the probability of rolling a two at least 3 times in 4 rolls. This can happen in two scenarios:
1. Rolling a two exactly 3 times.
2. Rolling a two exactly 4 times.

### Step 3: Calculate the probability for each scenario
**Scenario 1: Rolling a two exactly 3 times**

The number of ways to choose 3 rolls out of 4 to be twos is given by the binomial coefficient \( \binom{4}{3} \). The probability of getting a two 3 times and not getting a two 1 time is:
\[ P(\text{exactly 3 twos}) = \binom{4}{3} \left( \frac{1}{6} \right)^3 \left( \frac{5}{6} \right)^1 \]
Calculating this:
\[ \binom{4}{3} = 4 \]
Thus,
\[ P(\text{exactly 3 twos}) = 4 \left( \frac{1}{6} \right)^3 \left( \frac{5}{6} \right) = 4 \cdot \frac{1}{216} \cdot \frac{5}{6} = \frac{20}{1296} \]

**Scenario 2: Rolling a two exactly 4 times**

The probability of rolling a two 4 times is:
\[ P(\text{exactly 4 twos}) = \binom{4}{4} \left( \frac{1}{6} \right)^4 \left( \frac{5}{6} \right)^0 \]
Calculating this:
\[ \binom{4}{4} = 1 \]
Thus,
\[ P(\text{exactly 4 twos}) = 1 \left( \frac{1}{6} \right)^4 = \frac{1}{1296} \]

### Step 4: Combine the probabilities
Now, we add the probabilities from both scenarios to find the total probability of rolling a two at least 3 times:
\[ P(\text{at least 3 twos}) = P(\text{exactly 3 twos}) + P(\text{exactly 4 twos}) = \frac{20}{1296} + \frac{1}{1296} = \frac{21}{1296} \]

### Step 5: Simplify the probability
We can simplify \( \frac{21}{1296} \):
\[ \frac{21}{1296} = \frac{7}{432} \]

### Final Answer
The probability of rolling a two at least 3 times when rolling a die 4 times is:
\[ \frac{7}{432} \]