Question

a lock is opened using a sequence of three numbers. the numbers range from 0 to 39, inclusive, and cannot be repeated within the sequence. what is the probability that all the numbers in the sequence are even? express your answer as a percent and round to the nearest whole number. 8% 12% 16% 24%

Explanation:

Step1: Find total even numbers

Numbers from 0 to 39 inclusive. Even numbers: 0,2,4,...,38. The count is $\frac{39 - 0}{2}+ 1=20$ (since formula for number of terms in arithmetic sequence: $n=\frac{a_n - a_1}{d}+1$, here $a_1 = 0,a_n = 38,d = 2$). Total numbers: $40$ (from 0 to 39 inclusive).

Step2: Calculate total permutations (no repetition)

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

Step3: Calculate permutations of even numbers (no repetition)

We choose 3 non - repeated even numbers. The number of even numbers is 20. The number of ways to choose and arrange 3 even numbers from 20 is $P(20,3)=\frac{20!}{(20 - 3)!}=20\times19\times18$.

Step4: Calculate probability

Probability $P=\frac{\text{Number of favorable outcomes}}{\text{Number of total outcomes}}=\frac{20\times19\times18}{40\times39\times38}$.

Simplify the fraction:

$\frac{20\times19\times18}{40\times39\times38}=\frac{20}{40}\times\frac{19}{39}\times\frac{18}{38}=\frac{1}{2}\times\frac{19}{39}\times\frac{9}{19}=\frac{9}{78}\approx0.1154$.

Convert to percentage: $0.1154\times100\approx12\%$.

Answer:

B. 12%