If a three digit number is randomly chosen. What is the probability that either the number itself or some permutation of the number (which is a three digit number) is divisible by 4 and 5?
Solution in this problem we need to find the probaility of that all the three digit numbers which can be divisible by 4 and 5 for this we will have the following steps ,All the three digits number from 100 to 999
Divisibility by 4: its last 2 digits are divisible by 4.
Divisibility by 5: its last digit is 5 0r 0.
So, possible last two digits are: 20, 40, 60, 80, 00 = 5 ways.
Also, ways of selecting the other digit(at hundreds place) = 9 (1,2,3,…,9)
Hence, total favorable ways = 9 x 5 = 45
Also, total ways of selecting 3 digit numbers = 9 x 10 x 10 = 900 (1st number we will take because there should not be 0, because 011 means 2 digits not three digits so other place like tenth and hundredth can be 10 and 10.)
So, required probability =Favorable outcomes/total out comes= 45/900 = 1/20
So our answer will be 1/20 .
Solution in this problem we need to find the probaility of that all the three digit numbers which can be divisible by 4 and 5 for this we will have the following steps ,All the three digits number from 100 to 999
Divisibility by 4: its last 2 digits are divisible by 4.
Divisibility by 5: its last digit is 5 0r 0.
So, possible last two digits are: 20, 40, 60, 80, 00 = 5 ways.
Also, ways of selecting the other digit(at hundreds place) = 9 (1,2,3,…,9)
Hence, total favorable ways = 9 x 5 = 45
Also, total ways of selecting 3 digit numbers = 9 x 10 x 10 = 900 (1st number we will take because there should not be 0, because 011 means 2 digits not three digits so other place like tenth and hundredth can be 10 and 10.)
So, required probability =Favorable outcomes/total out comes= 45/900 = 1/20
So our answer will be 1/20 .
