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: $n=\frac{l - a}{d}+1$, here $a = 0$, $l=38$, $d = 2$). Total numbers: 40 (from 0 to 39 inclusive).

Step2: Calculate total permutations (no repetition)

Total number of ways to choose 3 numbers out of 40 without repetition (permutations) is $P(40,3)=\frac{40!}{(40 - 3)!}=40\times39\times38$.

Step3: Calculate permutations of even numbers

Number of ways to choose 3 even numbers out of 20 without repetition is $P(20,3)=\frac{20!}{(20 - 3)!}=20\times19\times18$.

Step4: Calculate probability

Probability $P=\frac{P(20,3)}{P(40,3)}=\frac{20\times19\times18}{40\times39\times38}$. Simplify: $\frac{20}{40}\times\frac{19}{39}\times\frac{18}{38}=\frac{1}{2}\times\frac{19}{39}\times\frac{9}{19}=\frac{9}{78}\approx0.1154$. Convert to percentage: $0.1154\times100\approx12\%$.

Answer:

12% (Option: 12%)