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

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Mathematics for JEE Mains/Advanced, XI & XII (All Boards)

Since the numbers selected randomly can be repeated hence ans is 16^3/10^3
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A number divisible by 2 AND 3 is divisible by 6. 100/6 = 16.67 There are 16 numbers between 1 and 100 which are divisible by 6. Out of these 16 we have to select 3 numbers. The number of ways to do this is 16C3. Total sample space is selection of 3 numbers from 100 numbers. The no. of ways to...
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A number divisible by 2 AND 3 is divisible by 6. 100/6 = 16.67 There are 16 numbers between 1 and 100 which are divisible by 6. Out of these 16 we have to select 3 numbers. The number of ways to do this is 16C3. Total sample space is selection of 3 numbers from 100 numbers. The no. of ways to do this is 100C3. So, probability of selecting 3 numbers, which are divisible by 2 and 3 i.e by 6 out of 1st 100 natural numbers is = 16C3/100C3 = 16*15*14/(100*99*98) = 0.0035 (approx... rounded off). read less
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Abhishek S ( M.Pharm.MBA) tutor for IB/International/IGCSE

Divisible by 2= 50, by 3= 33, by 6= 16 so 16/100
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Headstart to assured 95+ score in math

(16x15x14) / (100x99x98)
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Headstart to assured 95+ score in math

(16x15x14) / (100x99x98). hope its correct :)
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IIT Delhi Final Year Student

numbers divisible by 2 and 3 are multiple of 6. there are 16 numbers divisible by 6 till 1-100. so probability is 16C3/100C3
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Online Tutor-Mathematics, Science, Competitive Exam Preparation

To be divisible by both 2&3 the numbers should be multiples of 6. There are 16 such numbers 6,12,18,....,96. To choose 3 out of these 16 is 16c3. This comes in numerator as it is a possible outcome. Total possible outcome is 100c3. So answer is (16c3)/(100c3).
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The probability of three natural nos to be selected from 100 nos randomly is 3/100. Moreover there are 16 natural nos that are both divisible by 2 & 3 starting from 6 & ending at 96.Therefore, required prob = 3 X 16/100.
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Solution: Numbers divisible by 2 and 3 also divisible by 6. so, Divisors of 6 between 1 to 100 = 6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96= 16 numbers Probability of selecting 3 natural numbers = 16C3/ 100C3 =(16x15x14)/(100x99x98) simplify it further..
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AMIE

S = { 1, 2 3, ............, 100 } E = { 6, 12, 18, 24, .....96 } P(E) = 16C3 / 100C3 16 x 15 x 14 3 x 2 x 1 = ------------------ x ------------------ 3 x 2 x 1 100 x 99 x 98 4 x 5 = --------------- ...
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S = { 1, 2 3, ............, 100 } E = { 6, 12, 18, 24, .....96 } P(E) = 16C3 / 100C3 16 x 15 x 14 3 x 2 x 1 = ------------------ x ------------------ 3 x 2 x 1 100 x 99 x 98 4 x 5 = --------------- 25 x 33 4 = ---------- 165 read less
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