Math  /  Discrete

Question2. Let AA be the set of all strings of decimal digits of length 6 .
For example, 010312 and 194483 are two strings in A. One string from AA is selected at random. Find the probability that: a. The string begins with 0123 . b. The string contains no digit greater than 6 . c. The string contains at least one digit greater than 6 ? d. The string has at least one digit that repeats?

Studdy Solution

STEP 1

What is this asking? We're picking a random 6-digit string.
What's the chance it starts with "0123", has only digits 0-6, has at least one digit bigger than 6, and has at least one repeated digit? Watch out! There are 10 possible digits (0 through 9), and the strings can start with a zero!
Don't forget about the complement rule for "at least one".

STEP 2

1. Calculate the total number of possible strings.
2. Calculate the probability of starting with "0123".
3. Calculate the probability of containing no digit greater than 6.
4. Calculate the probability of containing at least one digit greater than 6.
5. Calculate the probability of having at least one repeated digit.

STEP 3

We have a 6-digit string, and each digit can be any of the 10 digits from 0 to 9.
So, for each digit, we have 10 choices.

STEP 4

Since there are 6 digits, the **total number of possible strings** is 101010101010=106=1,000,00010 \cdot 10 \cdot 10 \cdot 10 \cdot 10 \cdot 10 = 10^6 = \mathbf{1,000,000}.
One million possibilities!

STEP 5

If the string starts with "0123", the first four digits are fixed.
We only get to choose the last two digits freely.

STEP 6

Each of the last two digits can be any of the 10 digits.
So, there are 1010=102=10010 \cdot 10 = 10^2 = 100 such strings.

STEP 7

The **probability** of starting with "0123" is the number of strings starting with "0123" divided by the total number of strings: 1001,000,000=110,000=0.0001\frac{100}{1,000,000} = \frac{1}{10,000} = \mathbf{0.0001}.

STEP 8

If no digit is greater than 6, each digit can be one of the 7 digits from 0 to 6.

STEP 9

Since there are 6 digits in the string, there are 777777=76=117,6497 \cdot 7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 = 7^6 = \mathbf{117,649} such strings.

STEP 10

The **probability** of having no digit greater than 6 is 117,6491,000,000=0.117649\frac{117,649}{1,000,000} = \mathbf{0.117649}.

STEP 11

This is the complement of the previous case!
It's easier to calculate 1 minus the probability of *no* digits greater than 6.

STEP 12

So, the **probability** of at least one digit greater than 6 is 10.117649=0.8823511 - 0.117649 = \mathbf{0.882351}.
Much easier than counting all the possibilities directly!

STEP 13

Again, it's easier to think about the complement: the probability of having *no* repeated digits.

STEP 14

For no repeated digits, the first digit can be any of the 10 digits.
The second digit can be any of the remaining 9 digits, the third any of the remaining 8, and so on.

STEP 15

So, there are 1098765=151,20010 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 = \mathbf{151,200} strings with no repeated digits.

STEP 16

The probability of no repeated digits is 151,2001,000,000=0.1512\frac{151,200}{1,000,000} = 0.1512.

STEP 17

Therefore, the **probability** of at least one repeated digit is 10.1512=0.84881 - 0.1512 = \mathbf{0.8488}.

STEP 18

a. The probability the string begins with 0123 is 0.0001. b. The probability the string contains no digit greater than 6 is 0.117649. c. The probability the string contains at least one digit greater than 6 is 0.882351. d. The probability the string has at least one repeated digit is 0.8488.

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