+1124 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..
7426 , 7428 , 7624 , 7628 , 7824 , 7826 , 9426 , 9428 , 9624 , 9628 , 9824 , 9826 , Total = 12 permutations.
Probability ==1 / 12 ==0.08333....
+1124 Hi guest,
So this would be the elegant way of doing it?...writing each combination out?..Thanks for your response
+1124 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
, ,
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.
+1124 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
\((5*5*1*1)+(4*5*1*1)+(4*5*1*1)\)
\(=25+20+20\)
\(=65\)
\(P = {65 \over 840}={13 \over 168}=0,08\)
