1. **Problem statement:** We have a binomial random variable $X \sim \text{Binomial}(n=5, p=0.55)$ representing the number of cancer patients (out of 5) with WBC < 3000. 2. We want to find the probability that no more than 50% of the five patients have WBC < 3000, i.e., $P(X \leq 2)$ since 50% of 5 is 2.5 and "no more than 50%" means at most 2 patients. 3. The binomial probability formula is: $$ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} $$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. 4. To find $P(X \leq 2)$, sum probabilities for $k=0,1,2$: $$ P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) $$ 5. Calculate each term: - $P(X=0) = \binom{5}{0} (0.55)^0 (0.45)^5 = 1 \times 1 \times 0.45^5 = 0.0185$ - $P(X=1) = \binom{5}{1} (0.55)^1 (0.45)^4 = 5 \times 0.55 \times 0.0410 = 0.1128$ - $P(X=2) = \binom{5}{2} (0.55)^2 (0.45)^3 = 10 \times 0.3025 \times 0.0911 = 0.2757$ 6. Sum these probabilities: $$ P(X \leq 2) = 0.0185 + 0.1128 + 0.2757 = 0.4069 $$ 7. **Final answer:** The probability that no more than 50% of the five patients have WBC < 3000 is approximately **0.407**.