8. (Devore : Section 2.3 #41) An ATM personal identification number (PIN) consis

Question

8. (Devore : Section 2.3 #41) An ATM personal identification number (PIN) consists of four digits, each a 0, 1,2,. ..,8 or 9, in succession. (a) How many different possible PINs are there if there are no restrictions on the choice of digits? (b) According to a representative at the author's local branch of Chase Bank, there are in fact restrictions on the choice of digits. The following choices are prohibited: (i) all four digits identical (i) sequences of consecutive ascending or descending digits, such as 6543 (ii) any sequence starting with 19 (birth years are too easy to guess). So if one of the PINs in (a) is randomly selected, what is the probability that it will be a legitimate PIN (that is, not be one of the prohibited sequences)? (c) Someone has stolen an ATM card and knows that the first and last digits of the PIN are 8 and 1, respectively. He has three tries before the card is retained by the ATM (but does not realize that). So he randomly selects the 2nd and 3rd digits for the first try, then randomly selects a different pair of digits for the second try, and yet another randomly selected pair of digits for the third try (the individual knows about the restrictions described in (b) so selects only from the legitimate possibilities). What is the probability that the individual gains access to the account? (d) Recalculate the probability in (e) if the first and last digits are 1 and 1, respectively.

Explanation / Answer

(a) There are no restrictions on the choice of digits for the PIN. So, we have 10 options (0 to 9) to choose from, for each position of the PIN.

Hence, the number of possible combinations = 10x10x10x10 = 10000

(b)

(i) possible Combinations where all four digits are equal = 10 (e. g. 0000,1114,2222,...9999).

(ii) Number of possible combinations with consecutive decending digits = 7. (9876,8765,7654,...3210)

Similarly, possible combinations with consecutive ascending digits =7

(iii) possible Combinations with starting digits 19 = 10x10 (since we have 10 digits to choose from, for each of the vacant positions.

So total number of combinations prohibited = 10+14+100=124

The probability that a randomly selected combination of 4 digits is a legitimate PIN = (10000-124)/10000=0.9876

(c) with the first and last digits fixed, there are 100 possible combinations for the middle 2 digits.

So the probability that the thief gets the combination right in the first attempt = 1/100 = 0.01

With the first combination wrong, there are 99 possible combinations left. Probability that the thief gets the combination right in the 2nd attempt = probability of the first combination being wrong x probability that the second combination is right = 0.99 x (1/99) = 0.01

Similarly, probability of getting the combination right in the 3rd attempt = probability of first combination wrong x probability that the second combination wrong x probability that the 3rd combination right = (99/100) x (98/99) x (1/98) = 0.01

So the required probability = 0.01+0.01+0.01=0.03

(d) this time, there are 89 possibilities in all, since we have to exclude sequences starting with 19 and the sequence 1111.

So, on similar lines as in (c), the required probability = 1/89+ (88/89)x(1/88) + (88/89) x(87/88)x (1/87) = 3/89 = 0.0337

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