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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > If three natural numbers from 1 to 100 are selected...
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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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Divisible by 2= 50, by 3= 33, by 6= 16 so 16/100
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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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Data Analytics & Advanced Statistics Tutor with 18 years of experience in teaching

Total numbers that are divisible by both 2 & 3 from 1 to 100 natural numbers = 16.So required probability = 16C3/100C3
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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 = --------------- ...
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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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16C3/100C3
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Assuming you understand the basics of probability, Here sample space is 1, 2, 3, 4, ..., 99, 100 i.e. S = {1, 2, 3, 4, ..., 99, 100} so n(S) = 100 Here Event is set of numbers divisible by 2 and 3 (or we can say divisible by 6) And Event E = {6, 12, 18, 24, ...., 90, 96} i.e. n(E) = 16 So...
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Assuming you understand the basics of probability, Here sample space is 1, 2, 3, 4, ..., 99, 100 i.e. S = {1, 2, 3, 4, ..., 99, 100} so n(S) = 100 Here Event is set of numbers divisible by 2 and 3 (or we can say divisible by 6) And Event E = {6, 12, 18, 24, ...., 90, 96} i.e. n(E) = 16 So probability of this event will be P(E) = n(E)/n(S) = 16/100 = 4/25 (or 0.16) read less
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16C3/100C3
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The number which can be divisible by 2 as well as 3 will be like (2*3)k. So any number which is multiple of 6 will get divisible by 6. From 1 to 100 there are 16 numbers which are divisible by 6. So number of ways to choose 3 from 16 is 16C3. Probability will (16C3)/100C3
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Count the events divide them by 100.
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