If three natural numbers from 1 to 100 are selected randomly , then find the probability that all are divisible by both 2 and 3 ?

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Sabyasachi Dey

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A number is divisible by both 2 and 3 if it is divisible by 6. There are 16 such numbers in 1 to 100.(6,12,18,24....,96). So all my numbers should be from these 16 numbers, which can happen in 16c3 ways. But total number of possible choice is 100c3. So probability is 16c3/100c3
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There are 16 numbers divisible by 6 Thus probability that 3 numbers are divisible by both 2 and 3 = 16C3/100C3 = (16*15*14)/(100*99*98) = 4/(5*33*7) = 4/1155 = 0.0035
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Math magician

Shailendar's answer is correct
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Teaching is my passion!!!!

16/100
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GMAT Math Expert

if a number has to be divisible by both 2 and 3 then it should be divisible by 6 (LCM(2,3)=6). Now all multiples of 6 from 1 to 100 are divisible by both 2 and 3 and there are16 multiples of 6. Therefor probability is 16/100
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Expert tutor - Target Bank PO 2015 here

Sample Space will be multiple of 6 because numbers divided by 6 are the only numbers divided by 2 and 3. So Multiple of 6 below 100 are {6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96} =16 numbers Probability = 16/100 If it has 2 or 3 then it would have been different case :)
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1538415

4/1155
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Probability that the number is divisible by both 2 and 3== proability that the number is divisible by 6=> # of numbers divisible by 6 and < 100 = 16 Probability = (16C3)/(100C3)
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Dedicated Teacher With Lots of Teaching Experience In Maths For All Boards

16/100
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Math Educator for Std.11th ,12th , Engineering Entrance and Degree Level with 11+ Years Experience

Probability that all are divisible by both 2 and 3 = 4 / 1155
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