Question

a lock is opened using a sequence of three numbers. the numbers range from 0 to 29, 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% 24%

Explanation:

Step1: Calculate total number of possible sequences

The total number of ways to choose a sequence of 3 non - repeated numbers from 30 numbers (0 - 29) is given by the permutation formula $P(n,r)=\frac{n!}{(n - r)!}$, where $n = 30$ and $r=3$. So $P(30,3)=\frac{30!}{(30 - 3)!}=\frac{30!}{27!}=30\times29\times28 = 24360$.

Step2: Calculate number of sequences with all even numbers

There are 15 even numbers from 0 to 29 (0, 2, 4, …, 28). The number of ways to choose a sequence of 3 non - repeated even numbers is $P(15,3)=\frac{15!}{(15 - 3)!}=\frac{15!}{12!}=15\times14\times13=2730$.

Step3: Calculate the probability

The probability $P$ that all the numbers in the sequence are even is $P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{2730}{24360}\approx0.112$.

Step4: Convert to percentage and round

To convert the probability to a percentage, we multiply by 100: $0.112\times100 = 11.2\%$. Rounding to the nearest whole number, we get 11%. But since 11% is not in the options and the closest value among the given options to 11.2% is 12%.

Answer:

12%