a lock is opened using a sequence of three numbers. the numbers range from 0 to 39, inclusive, and cannot be…

Explanation:

Step1: Calculate total number of combinations

The total number of ways to choose 3 non - repeated numbers from 40 numbers (0 to 39) is given by the permutation formula $P(n,r)=\frac{n!}{(n - r)!}$, where $n = 40$ and $r=3$. So $P(40,3)=\frac{40!}{(40 - 3)!}=\frac{40!}{37!}=40\times39\times38=59280$.

Step2: Calculate number of combinations with all even numbers

There are 20 even numbers from 0 to 39. The number of ways to choose 3 non - repeated even numbers is $P(20,3)=\frac{20!}{(20 - 3)!}=\frac{20!}{17!}=20\times19\times18 = 6840$.

Step3: Calculate the probability

The probability $P$ that all the numbers in the sequence are even is $P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{6840}{59280}\approx0.1154$.

Step4: Convert to percentage and round

To convert to a percentage, we multiply by 100: $0.1154\times100 = 11.54%$. Rounding to the nearest whole number gives 12%.

Answer:

12%