Question 40

The probability that a 3-digit number formed from the digits 1—5 without repetition is divisible by 5 is 1/5.

To determine the probability, we note that a 3-digit number is divisible by 5 if its last digit is 5. Given the digits 1, 2, 3, 4, and 5, we can only use 5 in the last position for divisibility by 5, leading to a total of 12 valid combinations.

A) 1/10

This option suggests that the probability is 1 out of 10. However, considering the total number of 3-digit combinations formed from the digits 1—5 is 60, this fraction does not accurately reflect the scenario where only certain combinations are divisible by 5.

B) 1/5

This is the correct answer. The total number of 3-digit combinations formed from the digits 1—5 is 60. Only those combinations ending with the digit 5 are divisible by 5. There are 12 combinations that meet this criterion, leading to a probability of 12 out of 60, which simplifies to 1/5.

C) 1/4

This option implies that 1 out of 4 combinations would be divisible by 5. Given that only those ending in 5 qualify, and with only 12 qualifying out of 60 total combinations, this probability is inaccurate.

D) 2/5

This option suggests that 2 out of 5 combinations are divisible by 5. However, since only combinations ending in 5 are valid for divisibility, and there are only 12 such combinations, this estimate overstates the probability.

E) 1/2

This option posits that half of the 3-digit combinations are divisible by 5. This is incorrect as only 12 out of 60 combinations meet the divisibility condition, which is significantly lower than half.

Conclusion

The probability of a 3-digit number formed from the digits 1—5 without repetition being divisible by 5 is definitively 1/5, as only numbers ending with 5 satisfy the condition. All other options either overstate or misinterpret the relationship between the total combinations and those that meet the criteria for divisibility by 5.