(i) A die is thrown 7 times. If getting an "even number" is "success", find the probability of getting at least 6 successes.
(ii) A die is thrown 8 times. If getting an "even number" is a "success", find the probability of getting at least 7 successes.

Answer

To solve the given problems, we will use the binomial distribution formula. The binomial distribution is used when there are a fixed number of trials, each trial has two possible outcomes (success or failure), and the probability of success is constant.

### Part (i): A die is thrown 7 times. Find the probability of getting at least 6 successes.

1. **Identify Parameters**:
- Number of trials (n) = 7
- Probability of success (p) = Probability of getting an even number = 3/6 = 1/2
- Probability of failure (q) = 1 - p = 1/2

2. **Define Successes**:
- We need to find the probability of getting at least 6 successes, which means we need to find P(X ≥ 6).
- This can be calculated as P(X = 6) + P(X = 7).

3. **Calculate P(X = 6)**:
\[ P(X = 6) = \binom{7}{6} \left(\frac{1}{2}\right)^6 \left(\frac{1}{2}\right)^{7-6} = \binom{7}{6} \left(\frac{1}{2}\right)^7 \]
\[ = 7 \cdot \frac{1}{128} = \frac{7}{128} \]

4. **Calculate P(X = 7)**:
\[ P(X = 7) = \binom{7}{7} \left(\frac{1}{2}\right)^7 \left(\frac{1}{2}\right)^{0} = 1 \cdot \frac{1}{128} = \frac{1}{128} \]

5. **Combine Probabilities**:
\[ P(X \geq 6) = P(X = 6) + P(X = 7) = \frac{7}{128} + \frac{1}{128} = \frac{8}{128} = \frac{1}{16} \]

### Answer for Part (i):
The probability of getting at least 6 successes when a die is thrown 7 times is \( \frac{1}{16} \).

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### Part (ii): A die is thrown 8 times. Find the probability of getting at least 7 successes.

1. **Identify Parameters**:
- Number of trials (n) = 8
- Probability of success (p) = 3/6 = 1/2
- Probability of failure (q) = 1 - p = 1/2

2. **Define Successes**:
- We need to find the probability of getting at least 7 successes, which means we need to find P(X ≥ 7).
- This can be calculated as P(X = 7) + P(X = 8).

3. **Calculate P(X = 7)**:
\[ P(X = 7) = \binom{8}{7} \left(\frac{1}{2}\right)^7 \left(\frac{1}{2}\right)^{8-7} = \binom{8}{7} \left(\frac{1}{2}\right)^8 \]
\[ = 8 \cdot \frac{1}{256} = \frac{8}{256} = \frac{1}{32} \]

4. **Calculate P(X = 8)**:
\[ P(X = 8) = \binom{8}{8} \left(\frac{1}{2}\right)^8 \left(\frac{1}{2}\right)^{0} = 1 \cdot \frac{1}{256} = \frac{1}{256} \]

5. **Combine Probabilities**:
\[ P(X \geq 7) = P(X = 7) + P(X = 8) = \frac{1}{32} + \frac{1}{256} \]
To add these fractions, convert \( \frac{1}{32} \) to have a denominator of 256:
\[ \frac{1}{32} = \frac{8}{256} \]
So,
\[ P(X \geq 7) = \frac{8}{256} + \frac{1}{256} = \frac{9}{256} \]

### Answer for Part (ii):
The probability of getting at least 7 successes when a die is thrown 8 times is \( \frac{9}{256} \).

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