GATE CSE 2014 Set 2 | Question: 48

Question

The probability that a given positive integer lying between $1$ and $100$ (both inclusive) is NOT divisible by $2$, $3$ or $5$ is ______ .

Answer 1

Answer - $0.26$

Number of integers divisible by $2 = 50$

Number of integers divisible by $3 = 33$

Number of integers divisible by $5 = 20$

Number of integers divisible by $2$ and $3 = 16$

Number of integers divisible by $2$ and $5 = 10$

Number of integers divisible by $3$ and $5 = 6$

Number of integers divisible by $2$ and $3$ and $5 = 3$

Total numbers divisible by $2$ or $3$ or $5 = 50 + 33 + 20 -16 -10 - 6 + 3 = 74$

Total number not divisible by $2$ or $3$ or $5 = 26$

Probability $ = 0.26$ [EDIT]

Comments

option 2. ( Not div by 2) and ( not div by 3 ) and ( not div by 5)

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In case anyone has doubt whether to take $ceil$ or $floor$, remember we are finding number of numbers divisible by $x$ (say $3$) among a set of $100$ no.s. So $100/3 = 33.33$

If we take $34$ no.s then it’s wrong because we don’t have $34$ numbers. We actually have $33$ only. Hence $floor$ value is always taken.

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@Abhrajyoti00 can we conclude that whenever we want minimum we take ceil and when maximum asked then we take floor?

Answer 2

This is Brute Force method but takes very less time because we just need to check number is not div by 2,3 or 5.

Total no of possible outcomes N(s) = 100

N(e)=Number's not divisible by (2 OR 3 OR 5)  = (Not Div by 2  AND Not Div by 3 Not Div by 5 )      /// Demargon's law

N(e) = {1,7,11,13,17,19,23,29,31,37,41,43,47,49,53,59,61,67,71,73,77,79,83,89,91,97} = 26

Prob = N(e) / N(s) = 26/100 = 0.26

Answer is 0.26.

Comments

In examination setting,this takes long time and moreover, using this approach one may not be confident

Answer 3

There are total 100 numbers, out of which 

50 numbers are divisible by 2, 
33 numbers are divisible by 3,
20 numbers are divisible by 5

Following are counted twice above
16 numbers are divisible by both 2 and 3
10 numbers are divisible by both 2 and 5
6 numbers are divisible by both 3 and 5

Following is counted thrice above
3 numbers are divisible by all 2, 3 and 5

So total numbers divisible by 2, 3 and 5 are = = 50 + 33 + 20 - 16 - 10 - 6 + 3 = 103 - 29 = 74 So probability that a number is number is not divisible by 2, 3 and 5 = (100 - 74)/100 = 0.26

Answer 4

https://math.stackexchange.com/a/1034611

I think this one is a better way to find number of divisors. Please check and confirm if its good or not.