Question 1153493: A security code consists of a person's first and last initials and four digits. Use 8 decimals places.
(a) What is the probability of guessing a person's code on the first try?
(b) What is the probability of not guessing a person's code on the first try?
(c) You know a person's first name and that the last digit is odd. What is the probability of guessing this persons code?
(d) Are the statements in parts (a) - (c) examples of classical probability, empirical probability, or subjective probability?
Found 2 solutions by Boreal, jim_thompson5910:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! there are 26 choices for each of the first two and 10 for each digit
the number of possibilities is therefore 26^2*10000=6,760,000
the probability of guessing first try is 0.000000148
the probability of not is 1- that or 0.99999985
now 1*26*10*10*10*5 or 1 in 26000 or 0.00003846
classic probability, no evidence for actual experimental, and not subjective, since there is a specific rule.
Answer by jim_thompson5910(35256)  (Show Source):
You can put this solution on YOUR website!
(a)
Assuming you do not know the person's first or last name, this means there are 26 choices for each letter, making 26*26 = 676 choices so far.
For the digits portion, there are 10 choices per slot, and 4 slots, so this means 10^4 = 10,000 different combinations.
Overall, there are 676*10,000 = 6,760,000 different security codes.
There is only 1 correct security code, therefore 1/(6,760,000) = 0.000 000 147 929.
I have spaced out the decimal digits such that each group is composed of 3 digits. This is to hopefully make the number more readable.
Round that result to 8 decimal digits to get: 0.000 000 15
Answer: 0.000 000 15
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(b)
We'll use the result found in part (a)
P(not guessing code first try) = 1 - P(guessing code first try)
P(not guessing code first try) = 1 - 0.000 000 15
P(not guessing code first try) = 0.999 999 85
Answer: 0.999 999 85
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(c)
The first initial is unknown, so there are 26 letters to pick from. The last initial is known, so we have only one choice here. There are 26*1 = 26 different letter combos.
The first three digits are unknown, so we have 10 choices per slot and 3 slots, making 10^3 = 1,000 different combos so far. The last digit is odd. We have 5 choices here since we pick from this set {1,3,5,7,9}. There are 1,000*5 = 5,000 different combos of just the numbers alone.
Overall, there are 26*5,000 = 130,000 different security codes to try out if we know the person's first name and if we know the last digit is odd.
Divide 1 over this value:
1/(130,000) = 0.000 007 692 307 7
This rounds to 0.000 007 69
Answer: 0.000 007 69
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(d)
Answer: Classical probability
Classical probability is the idea of computing A/B
A = number of times a desired event occurs
B = total number of possible events
Specifically,
A = number of ways to get the correct code
B = number of codes possible
Empirical probability would not apply here because we don't have any actual experimental or empirical data to rely on. Subjective probability is not the answer either because the term "subjective" means it depends on the person doing the interpretation. It is better to go with objective data rather than subjective thoughts/beliefs/feelings.
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