Answer:
Probability of N Divisible by 3 - 0.33
Probability of N Divisible by 5 - 0.2
Probability of N Divisible by 7 - 0.413
Probability of N Divisible by 15 - 0.066
Probability of N Divisible by 105 - 0.0095
Step-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 = 333.3
Probability of N divisible by 3 {N%3 = 0 } ![= \frac{333.3}{1000} = 0.33](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B333.3%7D%7B1000%7D%20%3D%200.33)
total number of ways by which 1000 is divisible by 5 i.e. 1000/5 = 200
Probability of N divisible by 5 {N%5 = 0 } ![= \frac{200}{1000} = 0.2](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B200%7D%7B1000%7D%20%3D%200.2)
total number of ways by which 1000 is divisible by 7 i.e. 1000/7 = 142.857
Probability of N divisible by 7 {N%7 = 0 } ![= \frac{142.857}{1000} = 0.413](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B142.857%7D%7B1000%7D%20%3D%200.413)
total number of ways by which 1000 is divisible by 15 i.e. 1000/15 = 66.667
Probability of N divisible by 15 {N%15 = 0 } ![= \frac{66.667}{1000} = 0.066](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B66.667%7D%7B1000%7D%20%3D%200.066)
total number of ways by which 1000 is divisible by 105 i.e. 1000/105 = 9.52
Probability of N divisible by 105 {N%105 = 0 } ![= \frac{9.52}{1000} = 0.0095](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B9.52%7D%7B1000%7D%20%3D%200.0095)
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.