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The probability that a given positive integer lying between $1$ and $100$ (both inclusive) is NOT divisible by $2$, $3$ or $5$ is ______ .
asked in Probability by Veteran (106k points)
edited by | 1.7k views

3 Answers

+19 votes
Best answer
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]
answered by Loyal (9.1k points)
edited by
0
It is 100-74 = 26 rt?
0
yes my mistake i'll edit the answer
+4 votes
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

answered by Loyal (8.6k points)
+3 votes

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.
answered by Loyal (7k points)
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