Step 1: Think of it as summing many tiny probabilities.
Each one of the $3 \times 10^9$ base pairs in the haploid genome has an independent, very small chance, $5 \times 10^{-11}$, of picking up a mutation during a single replication. The expected number of mutations across the whole genome is just the sum of these individual per-base-pair probabilities, since expectation is additive even when the underlying events are rare and independent.
Step 2: Write the sum as a product.
Since every one of the $3 \times 10^9$ base pairs contributes the same small probability $5 \times 10^{-11}$ to the expected count, the sum of $3 \times 10^9$ identical terms is just that number of terms times one term:
\[ E[\text{mutations}] = (3 \times 10^{9}) \times (5 \times 10^{-11}) \]
Step 3: Simplify using powers of ten.
Group the plain numbers and the powers of ten separately:
\[ 3 \times 5 = 15, \qquad 10^{9} \times 10^{-11} = 10^{9 - 11} = 10^{-2} \]
\[ E[\text{mutations}] = 15 \times 10^{-2} \]
Converting $15 \times 10^{-2}$ to decimal form gives $0.15$, since dividing 15 by 100 shifts the decimal point two places left.
Step 4: Final answer.
So on average, about 0.15 mutations occur across the entire human haploid genome every time it is replicated, meaning roughly one mutation shows up every six or seven replications rather than every single one.
\[ \boxed{0.15} \]