Question

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 1

Explanation:

Step 1:

The variable N is to be selected at the random form which is {1,2,.....,1000}

the total number of way that 1000 divisible by 3 = 1000/3 = 333.3

Here the favorable case = 333 ways

The Exhaustive case = 1000 ways

Therefore, the probability that N can be divisible by 3 = 333/1000 = 0.333

Step 2:

The total number of way that 1000 divisible by 5 = 1000/5 = 200

Here the favorable case = 200 ways

Therefore, the probability that N can be divisible by 5 = 200/1000 = 0.2

Step 4:

The total number of ways that 1000 divisible by 7 = 1000/7 = 142.85 = 143

Here the favorable case = 143 ways

Step 5:

The total number of ways that 1000 divisible by 15 = 15/1000 = 66.6667 =67

Here the favorable case = 67 ways

Therefore, the probability that N can be divisible by 15 = 67/1000 = 0.067

Step 7:

Similarly, the probability of N can be divided by 105 = 9.5238/1000 = 0.0095

Last step:

than, Similarly N is selected from {1,2,.....(10)^k} where the variable k is larger than the probability that N can be:

Divisible by 3 = 10^k/3/10^k = 0.33

Divisible by 5 = 1/5 = 0.2

Divisible by 7 = 1/7 = 0.413

Divisible by 15 = 1/15 = 0.0666

Divisible by 105 = 1/105 = 0.0095

Answer 2

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

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 }
`= (333.3)/(1000) = 0.33`

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 }
`= (200)/(1000) = 0.2`

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 }
`= (142.857)/(1000) = 0.413`

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 }
`= (66.667)/(1000) = 0.066`

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 }
`= (9.52)/(1000) = 0.0095`

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.