# GATE2014-2-48

3.2k views
The probability that a given positive integer lying between $1$ and $100$ (both inclusive) is NOT divisible by $2$, $3$ or $5$ is ______ .

edited

Answer - $0.26$

no of integers divisible by $2 = 50$

no of integers divisible by $3 = 33$

no of integers divisible by $5 = 20$

no of integers divisible by $2$ and $3$ = $16$

no of integers divisible by $2$ and $5$ = $10$
no of integers divisible by $3$ and $5$ = $6$

no 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]

edited
0
It is 100-74 = 26 rt?
0
yes my mistake i'll edit the answer
1

NOT divisible by 2, 3 or 5

Does this mean it is

1. (not divisible by 2) AND (not divisible by 3) AND (not divisible by 5)

or does it mean

2. (not divisible by 2) OR (not divisible by 3) OR (not divisible by 5)

Can anybody explain why 2. is not the case? or is it?

1
What is the meaning of not divisible by 2,3 or 5?

1. (Not div by 2) or (not div by 3) or ( not div by 5)

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

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

0
In examination setting,this takes long time and moreover, using this approach one may not be confident

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.

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