A company found that at the end of each month, 2% of its one million data entries, contained errors. Today, one operator has made 49 computer entries.Calculate the probability that at most two (2) errors were made by this operator.

To solve this problem, we can use the binomial probability formula. The binomial probability formula is given by:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

where:

• $n$ is the number of trials (in this case, 49 entries),
• $k$ is the number of successes (in this case, errors),
• $p$ is the probability of success on a single trial (in this case, 2% or 0.02),
• $\binom{n}{k}$ is the binomial coefficient, calculated as $\frac{n!}{k!(n-k)!}$.

We need to find the probability that at most two errors were made, which means we need to calculate the probability for 0, 1, and 2 errors and then sum these probabilities.

1. Calculate the probability of 0 errors: $P(X = 0) = \binom{49}{0} (0.02)^0 (0.98)^{49}$ $P(X = 0) = 1 \cdot 1 \cdot (0.98)^{49}$ $P(X = 0) \approx (0.98)^{49} \approx 0.3642$

2. Calculate the probability of 1 error: $P(X = 1) = \binom{49}{1} (0.02)^1 (0.98)^{48}$ $P(X = 1) = 49 \cdot 0.02 \cdot (0.98)^{48}$ $P(X = 1) \approx 49 \cdot 0.02 \cdot 0.3715 \approx 0.3637$

3. Calculate the probability of 2 errors: $P(X = 2) = \binom{49}{2} (0.02)^2 (0.98)^{47}$ $P(X = 2) = \frac{49 \cdot 48}{2} \cdot (0.02)^2 \cdot (0.98)^{47}$ $P(X = 2) = 1176 \cdot 0.0004 \cdot 0.3791 \approx 0.1784$

Finally, sum the probabilities for 0, 1, and 2 errors: $P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$ $P(X \leq 2) \approx 0.3642 + 0.3637 + 0.1784 \approx 0.9063$

Therefore, the probability that at most two errors were made by this operator is approximately 0.9063, or 90.63%.