(a) Obtain binomial probability distribution, if :
(i) `n=6,p=(1)/(3)` (ii) `n=5,p=(1)/(6)`.
(b) Suppose X has a binomial distribution `B(6,(1)/(2))`.
Show that X = 3 is the most likely outcome.

Answer

To solve the problem, we will break it down into two parts as given in the question.

### Part (a): Obtain the binomial probability distribution

#### (i) For n = 6 and p = 1/3

1. **Identify q**:
\[ q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3} \]

2. **Write the binomial probability formula**:
The binomial probability formula is given by:
\[ P(X = x) = \binom{n}{x} p^x q^{n-x} \]
where \( n = 6 \), \( p = \frac{1}{3} \), and \( q = \frac{2}{3} \).

3. **Substitute values into the formula**:
\[ P(X = x) = \binom{6}{x} \left(\frac{1}{3}\right)^x \left(\frac{2}{3}\right)^{6-x} \]

4. **Calculate probabilities for x = 0 to 6**:
- For \( x = 0 \):
\[ P(X = 0) = \binom{6}{0} \left(\frac{1}{3}\right)^0 \left(\frac{2}{3}\right)^6 = 1 \cdot 1 \cdot \left(\frac{64}{729}\right) = \frac{64}{729} \]
- For \( x = 1 \):
\[ P(X = 1) = \binom{6}{1} \left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^5 = 6 \cdot \frac{1}{3} \cdot \frac{32}{243} = \frac{192}{729} \]
- For \( x = 2 \):
\[ P(X = 2) = \binom{6}{2} \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^4 = 15 \cdot \frac{1}{9} \cdot \frac{16}{81} = \frac{240}{729} \]
- For \( x = 3 \):
\[ P(X = 3) = \binom{6}{3} \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^3 = 20 \cdot \frac{1}{27} \cdot \frac{8}{27} = \frac{160}{729} \]
- For \( x = 4 \):
\[ P(X = 4) = \binom{6}{4} \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^2 = 15 \cdot \frac{1}{81} \cdot \frac{4}{9} = \frac{60}{729} \]
- For \( x = 5 \):
\[ P(X = 5) = \binom{6}{5} \left(\frac{1}{3}\right)^5 \left(\frac{2}{3}\right)^1 = 6 \cdot \frac{1}{243} \cdot \frac{2}{3} = \frac{12}{729} \]
- For \( x = 6 \):
\[ P(X = 6) = \binom{6}{6} \left(\frac{1}{3}\right)^6 \left(\frac{2}{3}\right)^0 = 1 \cdot \frac{1}{729} \cdot 1 = \frac{1}{729} \]

5. **Summarize the distribution**:
The binomial probability distribution for \( n = 6 \) and \( p = \frac{1}{3} \) is:
\[ P(X = 0) = \frac{64}{729}, \quad P(X = 1) = \frac{192}{729}, \quad P(X = 2) = \frac{240}{729}, \quad P(X = 3) = \frac{160}{729}, \quad P(X = 4) = \frac{60}{729}, \quad P(X = 5) = \frac{12}{729}, \quad P(X = 6) = \frac{1}{729} \]

#### (ii) For n = 5 and p = 1/6

1. **Identify q**:
\[ q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6} \]

2. **Write the binomial probability formula**:
\[ P(X = x) = \binom{5}{x} \left(\frac{1}{6}\right)^x \left(\frac{5}{6}\right)^{5-x} \]

3. **Calculate probabilities for x = 0 to 5**:
- For \( x = 0 \):
\[ P(X = 0) = \binom{5}{0} \left(\frac{1}{6}\right)^0 \left(\frac{5}{6}\right)^5 = 1 \cdot 1 \cdot \left(\frac{3125}{7776}\right) = \frac{3125}{7776} \]
- For \( x = 1 \):
\[ P(X = 1) = \binom{5}{1} \left(\frac{1}{6}\right)^1 \left(\frac{5}{6}\right)^4 = 5 \cdot \frac{1}{6} \cdot \frac{625}{1296} = \frac{3125}{7776} \]
- For \( x = 2 \):
\[ P(X = 2) = \binom{5}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^3 = 10 \cdot \frac{1}{36} \cdot \frac{125}{216} = \frac{1250}{7776} \]
- For \( x = 3 \):
\[ P(X = 3) = \binom{5}{3} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^2 = 10 \cdot \frac{1}{216} \cdot \frac{25}{36} = \frac{250}{7776} \]
- For \( x = 4 \):
\[ P(X = 4) = \binom{5}{4} \left(\frac{1}{6}\right)^4 \left(\frac{5}{6}\right)^1 = 5 \cdot \frac{1}{1296} \cdot \frac{5}{6} = \frac{25}{7776} \]
- For \( x = 5 \):
\[ P(X = 5) = \binom{5}{5} \left(\frac{1}{6}\right)^5 \left(\frac{5}{6}\right)^0 = 1 \cdot \frac{1}{7776} \cdot 1 = \frac{1}{7776} \]

4. **Summarize the distribution**:
The binomial probability distribution for \( n = 5 \) and \( p = \frac{1}{6} \) is:
\[ P(X = 0) = \frac{3125}{7776}, \quad P(X = 1) = \frac{3125}{7776}, \quad P(X = 2) = \frac{1250}{7776}, \quad P(X = 3) = \frac{250}{7776}, \quad P(X = 4) = \frac{25}{7776}, \quad P(X = 5) = \frac{1}{7776} \]

### Part (b): Show that X = 3 is the most likely outcome for X ~ B(6, 1/2)

1. **Identify parameters**:
Here, \( n = 6 \) and \( p = \frac{1}{2} \). Thus, \( q = 1 - p = \frac{1}{2} \).

2. **Write the binomial probability formula**:
\[ P(X = x) = \binom{6}{x} \left(\frac{1}{2}\right)^x \left(\frac{1}{2}\right)^{6-x} = \binom{6}{x} \left(\frac{1}{2}\right)^6 \]

3. **Calculate probabilities for x = 0 to 6**:
- For \( x = 0 \):
\[ P(X = 0) = \binom{6}{0} \left(\frac{1}{2}\right)^6 = 1 \cdot \frac{1}{64} = \frac{1}{64} \]
- For \( x = 1 \):
\[ P(X = 1) = \binom{6}{1} \left(\frac{1}{2}\right)^6 = 6 \cdot \frac{1}{64} = \frac{6}{64} = \frac{3}{32} \]
- For \( x = 2 \):
\[ P(X = 2) = \binom{6}{2} \left(\frac{1}{2}\right)^6 = 15 \cdot \frac{1}{64} = \frac{15}{64} \]
- For \( x = 3 \):
\[ P(X = 3) = \binom{6}{3} \left(\frac{1}{2}\right)^6 = 20 \cdot \frac{1}{64} = \frac{20}{64} = \frac{5}{16} \]
- For \( x = 4 \):
\[ P(X = 4) = \binom{6}{4} \left(\frac{1}{2}\right)^6 = 15 \cdot \frac{1}{64} = \frac{15}{64} \]
- For \( x = 5 \):
\[ P(X = 5) = \binom{6}{5} \left(\frac{1}{2}\right)^6 = 6 \cdot \frac{1}{64} = \frac{6}{64} = \frac{3}{32} \]
- For \( x = 6 \):
\[ P(X = 6) = \binom{6}{6} \left(\frac{1}{2}\right)^6 = 1 \cdot \frac{1}{64} = \frac{1}{64} \]

4. **Summarize the distribution**:
The probabilities are:
\[ P(X = 0) = \frac{1}{64}, \quad P(X = 1) = \frac{3}{32}, \quad P(X = 2) = \frac{15}{64}, \quad P(X = 3) = \frac{5}{16}, \quad P(X = 4) = \frac{15}{64}, \quad P(X = 5) = \frac{3}{32}, \quad P(X = 6) = \frac{1}{64} \]

5. **Identify the most likely outcome**:
The highest probability occurs at \( x = 3 \) where \( P(X = 3) = \frac{5}{16} \). Thus, \( X = 3 \) is the most likely outcome.