Step 1: Binomial Distribution Mean and Variance.
For $X \sim B(n,p)$, the mean is $\mu = np$ and the variance is $\sigma^2 = np(1-p)$.
Step 2: Apply Given Condition.
We are given $\mu + \sigma^2 = 1.8$. Substituting the formulas: $np + np(1-p) = np(2-p) = 1.8$. With $n=5$, this becomes: $5p(2-p) = 1.8$.
Step 3: Solve for $p$.
Expanding the equation: $10p - 5p^2 = 1.8$. Rearranging into a quadratic equation: $5p^2 - 10p + 1.8 = 0$. Dividing by 5: $p^2 - 2p + 0.36 = 0$. Using the quadratic formula: \[ p = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(0.36)}}{2(1)} = \frac{2 \pm \sqrt{4 - 1.44}}{2} = \frac{2 \pm \sqrt{2.56}}{2} = \frac{2 \pm 1.6}{2}. \] The possible values for $p$ are $p = \frac{2 + 1.6}{2} = 1.8$ or $p = \frac{2 - 1.6}{2} = 0.2$. Since $p$ must be between 0 and 1, we select $p = 0.2$.
Step 4: Calculate Probability of 2 Successes.
The probability of exactly 2 successes in 5 trials is given by the binomial probability formula: $P(X=2) = \binom{5}{2} (0.2)^2 (0.8)^{5-2}$. Calculating this: \[ P(X=2) = \binom{5}{2} (0.2)^2 (0.8)^3 = 10 \cdot 0.04 \cdot 0.512 = 0.2048. \]
Step 5: Final Result.
The probability of observing 2 successes is $0.2048$.