1. **Problem statement:** We have a binomial random variable $X \sim \text{Binomial}(n=5, p=0.55)$ and want to find the probability $P(X \leq 2)$, which means the probability that $X$ takes values 0, 1, or 2. 2. **Formula:** The binomial probability mass function is given by $$ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} $$ where $\binom{n}{k}$ is the binomial coefficient, $p$ is the probability of success, and $n$ is the number of trials. 3. **Calculate $P(X \leq 2)$:** $$ P(X \leq 2) = \sum_{k=0}^2 \binom{5}{k} (0.55)^k (0.45)^{5-k} $$ 4. **Calculate each term:** - For $k=0$: $$ \binom{5}{0} (0.55)^0 (0.45)^5 = 1 \times 1 \times 0.45^5 = 0.0185 $$ - For $k=1$: $$ \binom{5}{1} (0.55)^1 (0.45)^4 = 5 \times 0.55 \times 0.0410 = 0.1128 $$ - For $k=2$: $$ \binom{5}{2} (0.55)^2 (0.45)^3 = 10 \times 0.3025 \times 0.0911 = 0.2757 $$ 5. **Sum the probabilities:** $$ P(X \leq 2) = 0.0185 + 0.1128 + 0.2757 = 0.407 $$ 6. **Interpretation:** There is approximately a 40.7% chance that at most 2 successes occur in 5 trials with success probability 0.55 each.