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Suppose a 4-digit PIN is allowed to begin with any digit except zero and the remaining digits have no restriction. If repeated digits are allowed, find the probability of the PIN beginning with a digit greater than 6 and ending with a 5.

User Daff
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Final answer:

The probability of a 4-digit PIN beginning with a digit greater than 6 and ending with a 5, with no restrictions on the middle digits, is 1/3 or approximately 33.33%.

Step-by-step explanation:

To find the probability of a 4-digit PIN beginning with a digit greater than 6 and ending with 5, we must consider each position of the PIN independently, the first digit can be either 7, 8, or 9 by the given restriction, and there are 3 possibilities out of 9 acceptable starting digits (1-9). For the remaining two middle digits, there are 10 possibilities for each (0-9), since there are no restrictions, and the last digit must be a 5, so there is only 1 possibility out of 10.

Hence, you calculate the probability as follows:

  • First digit: 3 possibilities (7, 8, or 9)
  • Middle two digits: 10 possibilities each
  • Last digit: 1 possibility (must be 5)

The total number of 4-digit PINs starting with any digit except zero is 9*10*10*10 (since the first digit can be any from 1 to 9 and each of the remaining three digits can be any from 0 to 9). The number of favourable outcomes is 3*10*10*1. The probability is the number of favourable outcomes divided by the total number of possible outcomes:

Probability = (3*10*10*1) / (9*10*10*10) = 3/9 = 1/3.

Therefore, the probability of a 4-digit PIN starting with a digit greater than 6 and ending with a 5 is 1/3 or approximately 33.33%.

User Ryan Porter
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