Calculate P ( r ) for r = 1, 2, 3, 4 and 5 by using the recurrence formula of the binomial distribution for the following Hence, draw the histogram for the distribution :
(i) `p=(1)/(3),n=5` (ii) `p=(1)/(6),n=5`.

Answer

To solve the problem, we will calculate the probabilities \( P(r) \) for \( r = 1, 2, 3, 4, 5 \) using the recurrence formula of the binomial distribution for the two cases given: \( p = \frac{1}{3}, n = 5 \) and \( p = \frac{1}{6}, n = 5 \).

### Part (i): \( p = \frac{1}{3}, n = 5 \)

1. **Calculate \( P(1) \)**:
\[ P(1) = \binom{5}{1} \left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^{5-1} \]
\[ = 5 \cdot \frac{1}{3} \cdot \left(\frac{2}{3}\right)^4 = 5 \cdot \frac{1}{3} \cdot \frac{16}{81} = \frac{80}{243} \approx 0.329 \]

2. **Calculate \( P(2) \)**:
\[ P(2) = \binom{5}{2} \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^{5-2} \]
\[ = 10 \cdot \left(\frac{1}{3}\right)^2 \cdot \left(\frac{2}{3}\right)^3 = 10 \cdot \frac{1}{9} \cdot \frac{8}{27} = \frac{80}{243} \approx 0.329 \]

3. **Calculate \( P(3) \)**:
\[ P(3) = \binom{5}{3} \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^{5-3} \]
\[ = 10 \cdot \left(\frac{1}{3}\right)^3 \cdot \left(\frac{2}{3}\right)^2 = 10 \cdot \frac{1}{27} \cdot \frac{4}{9} = \frac{40}{243} \approx 0.165 \]

4. **Calculate \( P(4) \)**:
\[ P(4) = \binom{5}{4} \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^{5-4} \]
\[ = 5 \cdot \left(\frac{1}{3}\right)^4 \cdot \left(\frac{2}{3}\right)^1 = 5 \cdot \frac{1}{81} \cdot \frac{2}{3} = \frac{10}{243} \approx 0.041 \]

5. **Calculate \( P(5) \)**:
\[ P(5) = \binom{5}{5} \left(\frac{1}{3}\right)^5 \left(\frac{2}{3}\right)^{5-5} \]
\[ = 1 \cdot \left(\frac{1}{3}\right)^5 \cdot 1 = \frac{1}{243} \approx 0.004 \]

### Summary of Probabilities for \( p = \frac{1}{3}, n = 5 \):
- \( P(1) \approx 0.329 \)
- \( P(2) \approx 0.329 \)
- \( P(3) \approx 0.165 \)
- \( P(4) \approx 0.041 \)
- \( P(5) \approx 0.004 \)

### Part (ii): \( p = \frac{1}{6}, n = 5 \)

1. **Calculate \( P(1) \)**:
\[ P(1) = \binom{5}{1} \left(\frac{1}{6}\right)^1 \left(\frac{5}{6}\right)^{5-1} \]
\[ = 5 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^4 = 5 \cdot \frac{1}{6} \cdot \frac{625}{1296} = \frac{3125}{7776} \approx 0.402 \]

2. **Calculate \( P(2) \)**:
\[ P(2) = \binom{5}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{5-2} \]
\[ = 10 \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^3 = 10 \cdot \frac{1}{36} \cdot \frac{125}{216} = \frac{1250}{7776} \approx 0.161 \]

3. **Calculate \( P(3) \)**:
\[ P(3) = \binom{5}{3} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^{5-3} \]
\[ = 10 \cdot \left(\frac{1}{6}\right)^3 \cdot \left(\frac{5}{6}\right)^2 = 10 \cdot \frac{1}{216} \cdot \frac{25}{36} = \frac{250}{7776} \approx 0.032 \]

4. **Calculate \( P(4) \)**:
\[ P(4) = \binom{5}{4} \left(\frac{1}{6}\right)^4 \left(\frac{5}{6}\right)^{5-4} \]
\[ = 5 \cdot \left(\frac{1}{6}\right)^4 \cdot \left(\frac{5}{6}\right)^1 = 5 \cdot \frac{1}{1296} \cdot \frac{5}{6} = \frac{25}{7776} \approx 0.003 \]

5. **Calculate \( P(5) \)**:
\[ P(5) = \binom{5}{5} \left(\frac{1}{6}\right)^5 \left(\frac{5}{6}\right)^{5-5} \]
\[ = 1 \cdot \left(\frac{1}{6}\right)^5 \cdot 1 = \frac{1}{7776} \approx 0.0001 \]

### Summary of Probabilities for \( p = \frac{1}{6}, n = 5 \):
- \( P(1) \approx 0.402 \)
- \( P(2) \approx 0.161 \)
- \( P(3) \approx 0.032 \)
- \( P(4) \approx 0.003 \)
- \( P(5) \approx 0.0001 \)

### Histogram
To draw the histogram, plot the values of \( r \) on the x-axis (1 to 5) and the corresponding probabilities \( P(r) \) on the y-axis for both cases. Each bar will represent the probability of getting \( r \) successes in 5 trials.