Hello friends,

I trust you all had a great Easter!!...PLease would someone show me my error?..I'm going to abreviate the question a lot, as I'm sure everybody here has already seen something like this:.

a Four digit code - requiered - open lock.Must be even numbered. May not contain 0 or 1. No repeats.

Calculate the probability - open lock at first attempt - code > 5000 - 3rd digit is 2.

So I said:

For the 1st number, we can only have a 5,6,7,8,9. However, we need to remove an even, for the 4th number. That leaves a possibility of 1/4

For the 2nd number we may have 3,4,5,6,7,8,9. Remove an even number reserved for the 4rth number, and remove one extra for the number already used for the 1st number...That leaves a possibility of 1/5

The 3rd number is 2

The 4th number can be either 4,6,8, so this is a possibility of 1/3 OR remove 2 numbers for numbers already used in numbers1-2, which will give a probability of 1/1

Multiplying these outcomes together does not yield the answer of 0,08 as given in the answer sheet?...Please someone?..Thank you all..

juriemagic Apr 10, 2023

#1**0 **

7426 , 7428 , 7624 , 7628 , 7824 , 7826 , 9426 , 9428 , 9624 , 9628 , 9824 , 9826 , Total = 12 permutations.

Probability ==1 / 12 ==0.08333....

Guest Apr 10, 2023

#2**0 **

Hi guest,

So this would be the elegant way of doing it?...writing each combination out?..Thanks for your response

juriemagic
Apr 11, 2023

#5**0 **

I've done the hard work..there are 65 permutations. They are:

5324, 6524, 7826, 9428

5326, 6528, 7924, 9524

5328, 6724, 7926, 9526

5426, 6728, 7928, 9528

5428, 6824, 8324, 9624

5624, 6924, 8326, 9628

5628, 6928, 8426, 9724

5724, 7324, 8524, 9726

5726, 7326, 8526, 9728

5728, 7328, 8624, 9824

5824, 7426, 8724, 9826

5826, 7428, 8726

5924, 7524, 8924

5926, 7526, 8926

5928, 7528, 9324

6324, 7624, 9326

6328, 7628, 9328

6428, 7824, 9426

, ,

juriemagic Apr 11, 2023

#6**0 **

I think you should post your question here in the "answers" section exactly as it is in your book or assignment. There seems to be something missing from the info you have given. Don't worry about its length. Because initially I got 65 permutations as well, but that would to be out of a total of: 65 / (1/12) ==780 permutations, which seems not possible with the info you have given. Good luck.

Guest Apr 11, 2023

#8**0 **

I'll give the answer as they have it...

start with 5,7,9,or start wuith 6 or start with 8

\((3*5*1*3)+(1*5*1*2)+(1*5*1*2) \)

\(=45+10+10\)

\(=65\)

\(P = {65 \over 840}={13 \over 168}=0,08\)

OR

Ends in 4, or ends in 6 or end in 8

\((5*5*1*1)+(4*5*1*1)+(4*5*1*1)\)

\(=25+20+20\)

\(=65\)

\(P = {65 \over 840}={13 \over 168}=0,08\)

juriemagic
Apr 12, 2023