Question

5. Three digits are drawn at random and independently of each other, to generate one resulting 3-digit number. Each of the three digits can take the integer values 1 through 9, with each integer being equally likely to be chosen. That is, the resulting 3-digit number can take a value from 111 through 999 (but numbers that contain 0 are not possible). Calculate the following probabilities, rounding your answer to the nearest hundredth of a percent. (a) the chance that the 3-digit number is even ________%
 
 
 
 
 
 (b) the chance that the 3-digit number contains at least one digit that is a 1 ________%
 
 
 
 
 
 
 
 
 (c) the chance that all three digits in the 3-digit number are the same ________%
 
 
 


Answer 1

Total number of possible combinations out of three digits = 9*9*9 = 729

( As all the digits from 1 to 9 can take any place )

(a) the chance that the 3-digit number is even

The number will be even if the last digit is even

There are { 2, 4, 6, 8 } 4 even numbers

Total number of possible combinations for the number be even = 9*9*4

Total number of possible combinations for the number be even = 324

P[ the 3-digit number is even ] = 324/729

P[ the 3-digit number is even ] = 0.4444

P[ the 3-digit number is even ] = 44% ( rounded off )

(b) the chance that the 3-digit number contains at least one digit that is a 1

P[ At least on 1 ] =1 - P[ zero 1 ]

Number having zero 1 = 8*8*8 ( only 8 options )

Number having zero 1 = 512

P[ zero 1 ] = 512/729

P[ At least on 1 ] =1 - 512/729

P[ At least on 1 ] = 217/729

P[ At least on 1 ] = 0.2977

P[ At least on 1 ] = 30% ( rounded off )

(c) the chance that all three digits in the 3-digit number are the same

All the numbers are same, there are only 9 options { 111,222,333,444,555,666,777,888,999}

P[  all three digits in number are the same ] = 9/729

P[  all three digits in number are the same ] = 0.0123

P[  all three digits in number are the same ] = 1%