1. **State the problem:** We want to find the probability that the first digit $D$ in a data entry is greater than or equal to 2, i.e., $P(D \geq 2)$. 2. **Recall the probability distribution:** According to Benford's Law, the probabilities for each digit are given as: $$ P(1)=0.301, P(2)=0.176, P(3)=0.125, P(4)=0.097, P(5)=0.079, P(6)=0.067, P(7)=0.058, P(8)=0.051, P(9)=0.046 $$ 3. **Use the rule for probabilities:** Since $D$ can only be digits 1 through 9, and the total probability sums to 1, we can find $P(D \geq 2)$ by summing the probabilities for digits 2 through 9: $$ P(D \geq 2) = P(2) + P(3) + P(4) + P(5) + P(6) + P(7) + P(8) + P(9) $$ 4. **Calculate the sum:** $$ P(D \geq 2) = 0.176 + 0.125 + 0.097 + 0.079 + 0.067 + 0.058 + 0.051 + 0.046 $$ 5. **Perform the addition:** $$ P(D \geq 2) = 0.176 + 0.125 = 0.301 $$ $$ 0.301 + 0.097 = 0.398 $$ $$ 0.398 + 0.079 = 0.477 $$ $$ 0.477 + 0.067 = 0.544 $$ $$ 0.544 + 0.058 = 0.602 $$ $$ 0.602 + 0.051 = 0.653 $$ $$ 0.653 + 0.046 = 0.699 $$ 6. **Final answer:** $$ P(D \geq 2) = 0.699 $$ This means there is approximately a 69.9% chance that the first digit of a data entry is 2 or greater according to Benford's Law.