Randomly pick a number from the set containing numbers betwe

Randomly pick a number from the set containing numbers between 10000-99999 such that the digits are not repeated. You do not have to calculate the answer, just give the fraction you obtain. You are welcome to express the parts of the fraction in a product form. Justify your answer: (a) What is the probability that its first two digits are 16? (b) What is the probability that its last digit is 0? (c) What is the probability that its first digits are 16 or its last digit is 0?

Solution

Numbers in this set are all 5 digit numbers so that digits are not repeated

First digit has to be from 1 to 9. So 9 possibilities

Second digit cannot be same as first one but can be 0. So 9 possibilites

8 for third, 7 for fourth and 6 for fifth

So, 9^2*8*7*6=27216 numbers

a)

First two digits are 16. And non repeating digits. So next digit can have 8 possibilities, then 7 and then 6

So, 8*7*6 numbers

SO probability =(8*7*6)/( 9^2*8*7*6)=1/81

b)

Last digit is 0

So first digit can have 9 possibilites, next 8 ,next 7 and then 6

So, 9*8*7*6 numbers

Probability=(9*8*7*6)/( 9^2*8*7*6)=1/9

c)

Let, us count numbers so that first two digits are 16 and last digit is 0

So three digits are fixed. So 7 possibilities for third and 6 for fourth. So 7*6=42 numbers

So total numbers so that first two digits are 16 or last digit is 0

=Numbers with first digits 16+Numbers with last digit 0- Numbers with first digit 16 and last digit 0

= 8*7*6+9*8*7*6-7*6=7*6(8+9*8-1)=7*6*79

So probability =(7*6*79)/(9^2*8*7*6)=79/648