Question

1. Given a randomly chosen 4-digit number...

What is the probability...

a. That there are no pairs of consecutive digits in the number

b. That there are no pairs of consecutive digits in the number, if the last digit and first digit of the number could be classified as consecutive

c. Given a randomly chosen six digit number:

What is the probability that the number is an odd number if it contains the six digits 7-6-4-4-3-1, and is in a random order?

Answer 1

a.

The first digit can be chosen in 9 ways (Number 1 to  9)

The next digit can be any number except the consecutive digits (For eg, if the selected first digit is 1, then the second digit cannot be 2). So, the second digits can be chosen in 9 ways (Number 0 to  9, except the consecutive digits).

Similarly, the third and fourth digit can be chosen in 9 ways.

Possible numbers with no pairs of consecutive digits in the number = 9⁴ = 6561

Possible 4 digits numbers = 9 * 10 * 10 * 10 = 9000 (First digit can be Number 1 to  9 and last 3 digits can be any number from 0 to 9)

Probability of numbers with no pairs of consecutive digits in the number = 6561 / 9000 = 0.729

b.

The first digit can be chosen in 9 ways (Number 1 to  9)

The next digit can be any number except the consecutive digits (For eg, if the selected first digit is 1, then the second digit cannot be 2). So, the second digits can be chosen in 9 ways (Number 0 to  9, except the consecutive digits).

Similarly, the third digit can be chosen in 9 ways.

The fourth digit cannot be consecutive number of third digit and previous number of 1st digit. For eg if the first digit is 1 and third digit is 3. Then the fourth digit cannot be 4 or 0. Thus the fourth digit can be chosen in 8 ways.

Possible numbers with no pairs of consecutive digits in the number = 9 * 9 * 9 * 8 = 5832

Possible 4 digits numbers = 9 * 10 * 10 * 10 = 9000 (First digit can be Number 1 to  9 and last 3 digits can be any number from 0 to 9)

Probability of numbers with no pairs of consecutive digits in the number = 5832 / 9000 = 0.648

c.

Probability that the number is an odd number = Probability that the number contains 1, 3 or 7 in last digits and remaining numbers is first five digits.

Number of ways to choose last digits = 3

There are two 4's . Number of ways we can place two 4's in any of 5 places = ⁵C₂ = 10

Number of ways to choose remaining 3 digits = 3! = 3 * 2 * 1 = 6

Possible odd numbers = 3 * 10 * 6 = 180

For any number,

There are two 4's . Number of ways we can place two 4's in any of 6 places = ⁶C₂ = 15

Number of ways to choose remaining 4 digits = 4! = 4 * 3 * 2 * 1 = 24

Total possible numbers = 15 * 24 = 360

Probability that the number is an odd number = 180 / 360 = 0.5