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?

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 }

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 }

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 }

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 }

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 }

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.

d is the answer

step-by-step explanation: