# A credit card company randomly 9 generates temporary four digit passcodes for cardholders. Serena is expecting her credit card to arrive in the mail. What is the probability that her pass code will consist of four different odd digits?

First, determine your Sample Space ...

Since it was not stated in your problem whether or not the passcode digits must be unique, assuming that repetition is permitted (e.g., 3333)...

Next, ascertain Serena's occasion...

There are five odd numbers, but each one needs to be unique.

I hope that was useful.

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To calculate the probability of Serena's passcode consisting of four different odd digits, we first need to determine the total number of possible four-digit passcodes that can be generated with odd digits. There are 5 odd digits (1, 3, 5, 7, 9), and since the passcode consists of four digits, the total number of possible combinations of odd digits is (5 \times 5 \times 5 \times 5 = 625).

Next, we need to find the total number of possible four-digit passcodes regardless of whether they consist of odd or even digits. There are 10 digits in total (0 through 9), so the total number of possible combinations of four digits is (10 \times 10 \times 10 \times 10 = 10,000).

Therefore, the probability that Serena's passcode will consist of four different odd digits is (\frac{625}{10,000}), which simplifies to (0.0625) or (6.25%).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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