To solve the problem, we need to find the parameters of a binomial distribution given that the sum of the mean and variance for 6 trials is \( \frac{10}{3} \).
### Step-by-Step Solution:
1. **Identify the parameters of the binomial distribution**:
The mean \( \mu \) of a binomial distribution is given by:
\[
\mu = n \cdot p
\]
where \( n \) is the number of trials and \( p \) is the probability of success.
The variance \( \sigma^2 \) of a binomial distribution is given by:
\[
\sigma^2 = n \cdot p \cdot q
\]
where \( q = 1 - p \).
2. **Set up the equation**:
We are given that the sum of the mean and variance is:
\[
\mu + \sigma^2 = \frac{10}{3}
\]
Substituting the formulas for mean and variance into this equation gives:
\[
n \cdot p + n \cdot p \cdot q = \frac{10}{3}
\]
Since \( n = 6 \), we can substitute this in:
\[
6p + 6p(1 - p) = \frac{10}{3}
\]
3. **Simplify the equation**:
Simplifying the left-hand side:
\[
6p + 6p - 6p^2 = \frac{10}{3}
\]
This simplifies to:
\[
12p - 6p^2 = \frac{10}{3}
\]
4. **Multiply through by 3 to eliminate the fraction**:
\[
3(12p - 6p^2) = 10
\]
This gives:
\[
36p - 18p^2 = 10
\]
5. **Rearrange the equation**:
\[
-18p^2 + 36p - 10 = 0
\]
Dividing the entire equation by -2 for simplicity:
\[
9p^2 - 18p + 5 = 0
\]
6. **Use the quadratic formula to solve for \( p \)**:
The quadratic formula is given by:
\[
p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 9 \), \( b = -18 \), and \( c = 5 \):
\[
p = \frac{18 \pm \sqrt{(-18)^2 - 4 \cdot 9 \cdot 5}}{2 \cdot 9}
\]
\[
p = \frac{18 \pm \sqrt{324 - 180}}{18}
\]
\[
p = \frac{18 \pm \sqrt{144}}{18}
\]
\[
p = \frac{18 \pm 12}{18}
\]
This gives two possible values for \( p \):
\[
p = \frac{30}{18} = \frac{5}{3} \quad \text{(not valid since } p \leq 1\text{)}
\]
\[
p = \frac{6}{18} = \frac{1}{3}
\]
7. **Find \( q \)**:
Since \( q = 1 - p \):
\[
q = 1 - \frac{1}{3} = \frac{2}{3}
\]
8. **Write the binomial distribution**:
The binomial distribution can be expressed as:
\[
P(X = r) = \binom{n}{r} p^r q^{n-r}
\]
Substituting \( n = 6 \), \( p = \frac{1}{3} \), and \( q = \frac{2}{3} \):
\[
P(X = r) = \binom{6}{r} \left(\frac{1}{3}\right)^r \left(\frac{2}{3}\right)^{6-r}
\]
where \( r = 0, 1, 2, 3, 4, 5, 6 \).
### Final Answer:
The binomial distribution is given by:
\[
P(X = r) = \binom{6}{r} \left(\frac{1}{3}\right)^r \left(\frac{2}{3}\right)^{6-r}, \quad r = 0, 1, 2, 3, 4, 5, 6
\]
