Questions: A database system assigns a 32-character ID to each record, where each character is either a number from 0 to 9 or a letter from A to F. Assume that each number or letter being selected is equally likely. Find the probability that at least 20 characters in the ID are numbers. Use Excel to find the probability. - Round your answer to three decimal places.

Solution Steps

To solve this problem, we need to model it as a binomial distribution. Each character in the ID can be a number (0-9) or a letter (A-F), giving 16 possible characters. The probability of a character being a number is 10/16. We want to find the probability of at least 20 numbers in a 32-character ID.

1. Define the number of trials (n = 32) and the probability of success (p = 10/16).
2. Use the binomial distribution to calculate the probability of getting at least 20 numbers.
3. Sum the probabilities from 20 to 32.

We have a database system that assigns a 32-character ID, where each character can either be a number (0-9) or a letter (A-F). The total number of possible characters is 16. The probability of selecting a number is given by:

\[ p = \frac{10}{16} = 0.625 \]

The number of trials (characters in the ID) is:

\[ n = 32 \]

We need to find the probability of having at least 20 numbers in the ID. This can be expressed mathematically as:

\[ P(X \geq 20) = \sum_{k=20}^{32} P(X = k) \]

where \( P(X = k) \) follows a binomial distribution:

\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]

After calculating the probabilities for \( k \) from 20 to 32, we find:

\[ P(X \geq 20) \approx 0.5780802948421496 \]

Rounding this to three decimal places gives:

\[ P(X \geq 20) \approx 0.578 \]

Final Answer