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

Answer 1

Answer:

12

Step-by-step explanation:

Answer 2

Answer:

The probability is:

12%

Step-by-step explanation:

It is given that:

A lock is opened using a sequence of three numbers. The numbers range from 0 to 39.

Also, the numbers in the sequence could not repeat.

This means that there are total 20 even numbers.(0,2,4,6,.....,38)

and 20 odd numbers (i.e. 1,3,5,....,39)

Hence, the probability that all the numbers in the sequence are even is calculated as:

$\dfrac{20}{40}\times \dfrac{19}{39}\times \dfrac{18}{38}\\\\=\dfrac{3}{26}$

( Since, in the first place we have total 20 choices out of 40 numbers.

in the second place as one number has been chosen and there is no repetition hence we have 19 choices out of 39 numbers and similarly for the third place we have 18 choices out of the 38 available numbers)

Hence, in percentage the probability is given as:

$=\dfrac{3}{26}\times 100\\\\=11.5385\%$

to the nearest whole number it is 12%.

Hence, the probability that all the numbers in the sequence are even is:

12%