An integer N is to be selected at random from {1, 2, ... , (10)3 } in the sense that each integer has the same probability of being selected. What is the probability that N will be divisible by 3? by 5? by 7? by 15? by 105? How would your answer change if (10)3 is replaced by (lO)k as k became larger and larger?

Answer:Probability of N Divisible by 3 - 0.33Probability of N Divisible by 5 - 0.2Probability of N Divisible by 7 - 0.413Probability of N Divisible by 15 - 0.066Probability of N Divisible by 105 - 0.0095Step-by-step explanation:Given data:Integer N {1,2,.....10^3}Thus total number of ways by which 1000 is divisible by 3 i.e. 1000/3 = 333.3Probability of N divisible by 3 {N%3 = 0 } [tex]= \frac{333.3}{1000} = 0.33[/tex]total number of ways by which 1000 is divisible by 5 i.e. 1000/5 = 200Probability of N divisible by 5 {N%5 = 0 } [tex]= \frac{200}{1000} = 0.2[/tex]total number of ways by which 1000 is divisible by 7 i.e. 1000/7 = 142.857Probability of N divisible by 7 {N%7 = 0 } [tex]= \frac{142.857}{1000} = 0.413[/tex]total number of ways by which 1000 is divisible by 15 i.e. 1000/15 = 66.667Probability of N divisible by 15 {N%15 = 0 } [tex]= \frac{66.667}{1000} = 0.066[/tex]total number of ways by which 1000 is divisible by 105 i.e. 1000/105 = 9.52Probability of N divisible by 105 {N%105 = 0 } [tex]= \frac{9.52}{1000} = 0.0095[/tex]similarly for N is selected from 1,2.....(10)^k where K is  large then the N value. Therefore effect of k will remain same as previous part.