Question 1183743: I am confused on how to solve this, may you please help?
Use a factorial, permutation or combination to find the probability of the given event. Do NOT simplify your answer. Only consider digits greater than zero.
Bill forgot the last three digits of his 8 digit password. He only remembers that the digits are even and not the same. What is the probability he enters the correct password on the first try?
Found 2 solutions by Theo, greenestamps:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! there is only one possible code.
the possible digits are 2,4,6,8.
can't be 0
can't be any odd number between 0 and 9.
the probability that the first digit is the right digit is 1/4.
the probability that the second digit is the right digit, assuming that the the first digit is correct, is 1/3.
the probability that the third digit is the right digit, assuming that the first two digits are correct, is 1/2.
the total probability is 1/4 * 1/3 * 1/2 = 1/24.
if you look at the number of permutations possible, you will get 4P3 = 4! / (1!) = 4*3*2*1 / 1 = 24.
only one of those permutations is the correct permutation, therefore the probability is 1/24.
the permutation formula is p(n,x) = n! / (n-x)!.
when n = 4 and x = 3, the formula becomes p(4,3) = 4! / (4-3)! = 4! / 1! = 24.
only one permutation out of those 24 permutations is the correct permutation, therefore the probability is 1/24.
i'm pretty sure this is correct.
i was able to get the same answer both ways.
the denominator is the permutation of 4 possible numbers taken 3 at a time.
the numerator is 1.
the possible number are 2,4,6,8.
the possible 3 digit permutations are:
2 4 6 *****
2 6 4
4 2 6
4 6 2
6 2 4
6 4 2
2 4 8 *****
2 8 4
4 2 8
4 8 2
8 2 4
8 4 2
2 6 8 *****
2 8 6
6 2 8
6 8 2
8 2 6
8 6 2
4 6 8 *****
4 8 6
6 4 8
6 8 4
8 4 6
8 6 4
note that you have 4 sets of 6 = 24.
each set of 6 is a separate combination, if you don't take order of the digits into account
the permutation formula is p(n,x) = n! / (n-x)!
the combination formula is c(n,x) = n! / (x! * (n-x)!)
when n = 4 and x = 3, these formulas become:
p(4,3) = 4! / 3! = 4! = 24.
c(4,3) = 4! / (3! * 1!) = 4! / 3! = 4.
you have 4 possible combinatins where order is not important.
they are:
2 4 6
2 4 8
2 6 8
4 6 8
within each of those you have 6 possible arrangements where order is important.
4 * 6 = 24.
he will need to try up to 24 possible permutations and is assured to get the right combination.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The other tutor goes through a lengthy explanation of why there are only 24 possible sequences (permutations) of the last three digits, so that the probability of getting the right password on the first attempt is 1/24.
That is quite possibly the answer the author of the problem wanted; however, grammatically it is not the right answer.
The problem is with the phrasing of the question, which says the three last digits are "even and not the same".
That does NOT mean the same thing as "even and all different". For example, the digits 224 are all even and not the same....
So with the way the problem is stated, the only sequences NOT allowed are 222, 444, 666, and 888.
Without restrictions, the number of possible permutations of the last three digits is 4*4*4=64; with the 4 excluded ones, the number of allowable permutations is 64-4=60. And of course only one of them is the correct one.
ANSWER: The probability of entering the correct password on the first try is 1/60.
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