For the binomial distribution X with parameters `n=12,p=(1)/(2)`, compute
i) `P(10leXle12)" "`ii) `P(3 lt X lt 6)`

Answer

To solve the problem, we will compute the probabilities for the binomial distribution \( X \) with parameters \( n = 12 \) and \( p = \frac{1}{2} \).

### i) Compute \( P(10 \leq X \leq 12) \)

1. **Identify the Probability Formula**:
The probability mass function for a binomial distribution is given by:
\[ P(X = r) = \binom{n}{r} p^r (1-p)^{n-r} \]
Here, \( n = 12 \) and \( p = \frac{1}{2} \), so \( 1 - p = \frac{1}{2} \).

2. **Calculate \( P(X = 10) \), \( P(X = 11) \), and \( P(X = 12) \)**:
\[ P(X = 10) = \binom{12}{10} \left(\frac{1}{2}\right)^{10} \left(\frac{1}{2}\right)^{2} = \binom{12}{10} \left(\frac{1}{2}\right)^{12} \]
\[ P(X = 11) = \binom{12}{11} \left(\frac{1}{2}\right)^{11} \left(\frac{1}{2}\right)^{1} = \binom{12}{11} \left(\frac{1}{2}\right)^{12} \]
\[ P(X = 12) = \binom{12}{12} \left(\frac{1}{2}\right)^{12} = \binom{12}{12} \left(\frac{1}{2}\right)^{12} \]

3. **Calculate the Binomial Coefficients**:
\[ \binom{12}{10} = \binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66 \]
\[ \binom{12}{11} = 12 \]
\[ \binom{12}{12} = 1 \]

4. **Combine the Probabilities**:
\[ P(10 \leq X \leq 12) = P(X = 10) + P(X = 11) + P(X = 12) \]
\[ P(10 \leq X \leq 12) = \left(66 + 12 + 1\right) \left(\frac{1}{2}\right)^{12} = 79 \left(\frac{1}{2}\right)^{12} \]
\[ P(10 \leq X \leq 12) = \frac{79}{4096} \]

### ii) Compute \( P(3 < X < 6) \)

1. **Identify the Relevant Probabilities**:
We need to calculate \( P(X = 4) \) and \( P(X = 5) \):
\[ P(X = 4) = \binom{12}{4} \left(\frac{1}{2}\right)^{12} \]
\[ P(X = 5) = \binom{12}{5} \left(\frac{1}{2}\right)^{12} \]

2. **Calculate the Binomial Coefficients**:
\[ \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495 \]
\[ \binom{12}{5} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792 \]

3. **Combine the Probabilities**:
\[ P(3 < X < 6) = P(X = 4) + P(X = 5) \]
\[ P(3 < X < 6) = \left(495 + 792\right) \left(\frac{1}{2}\right)^{12} = 1287 \left(\frac{1}{2}\right)^{12} \]
\[ P(3 < X < 6) = \frac{1287}{4096} \]

### Final Answers:
- \( P(10 \leq X \leq 12) = \frac{79}{4096} \)
- \( P(3 < X < 6) = \frac{1287}{4096} \)