A bank issues a unique six-digit password to each of its online customers - Leaving Cert Mathematics - Question 1 - 2015

If we want to create a password that does not contain the digit zero, we can use any of the digits from 1 to 9, providing us with 9 possible values for each digit. Thus, the calculation for the total number of passwords that do not contain any zero is:

$$9^6 = 531441$$

Hence, there are 531,441 different passwords that do not contain any zero.

To find the probability that a randomly selected password contains at least one zero, we first calculate the total number of passwords containing zero and then use the complement rule.

The number of passwords that do not contain any zeros is already calculated as 531,441. Therefore, the total number of passwords that contain at least one zero is:

$$1,000,000 - 531441 = 468559$$

Now, the probability that a randomly selected password contains at least one zero is given by:

$$P( ext{at least one zero}) = \frac{468559}{1000000} = 0.468559$$

Thus, the probability is approximately 0.469.

To find the number of different ways to select three boxes from five available, we use the combination formula:

$$C(n, r) = \frac{n!}{r!(n-r)!}$$

Here, ( n = 5 ) and ( r = 3 ). The calculation becomes:

$$C(5, 3) = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10$$

Therefore, there are 10 different ways the bank can select the three required boxes.