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

Answer:

12%

Explanation:

Step1: Calculate number of available numbers

There are 40 numbers from 0 - 39. So, $n = 40$.

Step2: Calculate number of even numbers

Even numbers from 0 - 39 are 0, 2, ..., 38. There are 20 even numbers, so $m=20$.

Step3: Calculate probability of first - number being even

The probability that the first number in the sequence is even is $P_1=\frac{m}{n}=\frac{20}{40}=\frac{1}{2}$.

Step4: Calculate probability of second - number being even (without replacement)

After choosing the first even number, there are 19 even numbers left out of 39 remaining numbers. So $P_2=\frac{19}{39}$.

Step5: Calculate probability of third - number being even (without replacement)

After choosing the first two even numbers, there are 18 even numbers left out of 38 remaining numbers. So $P_3=\frac{18}{38}$.

Step6: Calculate overall probability

The probability that all three numbers are even is $P = P_1\times P_2\times P_3=\frac{20}{40}\times\frac{19}{39}\times\frac{18}{38}=\frac{20\times19\times18}{40\times39\times38}=\frac{6840}{59280}\approx 0.115$.

Step7: Convert to percentage and round

$0.115\times100 = 11.5%$, rounding to the nearest whole number gives 12%.