Question

Urgently plsss A randomly generated password contains four characters. Each of the four characters is either a lowercase letter or a digit from 0–9. Each character in the password cannot be used more than once.
What is the approximate probability that exactly one of the four characters will be a number?

1%
11%
28%
44%

Answer 1

Answer:

44%

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the values are is important(0123 is a different password of 3210), which means that the permutations formula is used to solve this question.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

$P_{(n,x)} = \frac{n!}{(n-x)!}$

Desired outcomes:

One number, from a set of 10

Three characters, from a set of 26.

They can be in 4 possible orders(CNNN, NCNN, NNCN, NNNC). So

$D = 4P_{10,1}P_{26,3} = 4*\frac{10!}{(10-9)!}*\frac{26!}{(26-3)!} = 624000$

Total outcomes:

Four characters, from a set of 26 + 10 = 36. So

$T = P_{36,4} = \frac{36!}{(36-4)!} = 1413720$

What is the approximate probability that exactly one of the four characters will be a number?

$P = \frac{D}{T} = \frac{624000}{1413720} = 0.4414$

So 44%.