Question

medal For Help ASAP :) PLEAAASSSE
The password for a school's grading system is a 4 digit code. If the password is composed of digits one through nine....
What is the probability that the password is composed of a one, a two, a three, and a four?
A.1/5,040
B.4/9
C.1/6,561
D.1/36
I think it's D ? Am I correct ?

Answer 1

@mathslover

Answer 2

@mtbender74 @Luigi0210

Answer 3

Can you digits repeat in the password?

Answer 4

i would assume, yes...

Answer 5

Just add up all the probabilities of each number being used. Like this

The chance of it being a 1 is 1/9
The chance of it being a 2 is 1/9
So far I have 2/9..

Just add them all up

Answer 6

So the number of codes that have 1, 2, 3, and 4 in any order would be 4*3*2*1

The total number of codes possible would be 9*9*9*9...

first number is numerator, second is denominator...

Answer 7

You have a total of 9^4 passwords possible. Out of these, the passwords using the digits 1, 2, 3 and 4 only are 4*3*2*1 = 24.

Hence probability of getting a password composed of the digits 1,2,3 and 4 is 24/9^4

Answer 8

Oh ok so it's b ?- I mulit them

Answer 9

From the answer choices it seems that the question is asking the probability of finding the password 1234 and not any password with these digits in any order. In this case, the probability is 1 over 9^4 = 1/6561 or choice C

Answer 10

@DemolisionWolf C or B ?

Answer 11

@mathslover c or B

Answer 12

If that is true @navk, i like this question less than when i first saw it :)

But i see what you mean...i hate poorly written questions.

Answer 13

@mtbender74 can it be B ?

Answer 14

I think it is C!

\(\cfrac{1}{9^4}\)

Answer 15

@navk i don't think so...i looked at that but that would include passcodes 1111, 2222, etc.

The problem says that the passcode has a 1, 2, 3, *and* 4...if that "and" said "or" i would agree with B

Answer 16

Then it's C for sure