If a license plate has 3 letters and then 3 numbers, what are the chances (probability?) of getting at least 1 letter "Z" in one of the letter spots if the letters are picked randomly?

and I think each letter can only be used once.

when you say, "at least 1 letter "Z" in one of the letter spots", you are implying that there could be 1,2, or 3 Z's in the letter spots.

But then you say that you think each letter can be used only once.

Usually on license plates, letters and numbers can be repeated, so I will work it out assuming that both letters and numbers can be repeated.

case 1: one Z
number of ways = 3x25x25x1000 = 1875000
case 2: two Z's
number of ways = 3x25x1000 = 75000
case 3: three Z's
number of ways = 1x1000 = 1000
total number of ways to have Z's = 1951000

number of ways without restrictions = 26x26x26x1000 = 17576000

so prob of at least 1 Z = 1951000/17576000 = ..111

other way:
no Z's at all = 25x25x25x1000 = 15625000
so number of ways for at least one Z
= 17576000 - 15625000 = 1951000
etc.

To calculate the probability of getting at least one "Z" in one of the letter spots on a license plate, we need to consider the number of favorable outcomes (plates with at least one "Z") and the total number of possible outcomes (all possible license plates).

Let's break down the calculation step by step:

Step 1: Determine the number of ways to arrange the letters.
Since each letter can only be used once and there are 3 spots for letters, the number of ways to arrange the letters is given by the permutation formula, denoted as P(n, r), which calculates the number of arrangements of r objects from a set of n distinct objects. In this case, we have 26 letters (excluding "Z") to choose from and 3 spots to fill. Therefore, the number of ways to arrange the letters is P(26, 3).

Step 2: Determine the number of ways to arrange the numbers.
Similarly, there are 10 numbers (0-9) to choose from and 3 spots to fill. So, the number of ways to arrange the numbers is P(10, 3).

Step 3: Calculate the total number of possible license plates.
Since the arrangement of letters and numbers are independent, we can multiply the number of ways to arrange the letters by the number of ways to arrange the numbers to get the total number of possible license plates. Therefore, the total number of license plates is P(26, 3) × P(10, 3).

Step 4: Determine the number of license plates that have at least one "Z".
To calculate this, we need to subtract the number of license plates with no "Z" from the total number of license plates. The number of license plates with no "Z" can be obtained by calculating the number of ways to arrange the remaining letters (25 letters excluding "Z") and the numbers. Thus, the number of license plates with at least one "Z" is P(25, 3) × P(10, 3).

Step 5: Calculate the probability.
Finally, to find the probability, divide the number of favorable outcomes (license plates with at least one "Z") by the total number of possible outcomes (all license plates). Therefore, the probability can be calculated as (P(25, 3) × P(10, 3)) / (P(26, 3) × P(10, 3)).

Simplifying the expression, we can cancel out the P(10, 3) terms, and the probability of getting at least one "Z" can be determined as P(25, 3) / P(26, 3).

You can plug in these values into a calculator or use mathematical software to compute the probability.