A 5-digit PIN code cannot begin with zero and the remaining digits have no restrictions. If repetition of digits is allowed, then what is the probability that the PIN code begins with a 7 and ends with an 8?
number of possible codes
= 9x10x10x10x10 = 90000
number starting with 7 and ending in 8
= 1x10x10x10x1 = 1000
prob(your event) = 1000/90000
= 1/90
A 4-digit pin code can begin with any number except 0,1 & 2. If repetition of the same
digit is allowed, the probability of a pin code begin with 5 and end with a 3 is?
To find the probability that the PIN code begins with a 7 and ends with an 8, we need to determine two things: the total number of possible PIN codes and the number of PIN codes that satisfy the given condition.
Total number of possible PIN codes:
Since the PIN code is 5 digits long and repetition of digits is allowed, each digit can take on any value from 0 to 9 (including 0). So, there are 10 possible digits for each position, giving us a total of 10^5 = 100,000 possible PIN codes.
Number of PIN codes that begin with a 7 and end with an 8:
Since the first digit cannot be zero, it can only be 7. Similarly, since the last digit has to be 8, it can only be 8. For the remaining 3 digits, there are no restrictions, so each digit can take on any value from 0 to 9. Thus, there are 10 possible digits for each of the 3 remaining positions.
Therefore, the number of PIN codes that begin with a 7 and end with an 8 is 10 * 10 * 10 = 1000.
Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = 1000 / 100,000
Probability = 0.01 or 1% (rounded to two decimal places)
Therefore, the probability that the PIN code begins with a 7 and ends with an 8 is 0.01 or 1%.
To find the probability that a 5-digit PIN code begins with a 7 and ends with an 8, we need to determine the total number of possible PIN codes and the number of PIN codes that meet the given criteria.
To start, let's determine the number of possible PIN codes. Since the PIN code cannot start with zero, we have 9 options for the first digit (1, 2, 3, 4, 5, 6, 7, 8, or 9). For the remaining four digits, there are no restrictions, so each digit has 10 options (0, 1, 2, 3, 4, 5, 6, 7, 8, or 9). Therefore, the total number of possible PIN codes is:
9 options for the first digit × 10 options for each of the remaining four digits = 9 × 10 × 10 × 10 × 10 = 90,000
Next, let's determine the number of PIN codes that begin with 7 and end with 8. Since the first digit must be 7 and the last digit must be 8, we still have no restrictions for the three remaining digits. Therefore, the number of PIN codes that meet the given criteria is:
1 option for the first digit (7) × 10 options for each of the three remaining digits × 1 option for the last digit (8) = 1 × 10 × 10 × 10 × 1 = 1,000
Finally, we can calculate the probability by dividing the number of PIN codes that meet the criteria by the total number of possible PIN codes:
Probability = Number of PIN codes that meet the criteria / Total number of possible PIN codes
= 1,000 / 90,000
= 1/90
≈ 0.0111 (rounded to four decimal places)
Therefore, the probability that a 5-digit PIN code begins with a 7 and ends with an 8 is approximately 0.0111 or 1/90.