Latest TCS Aptitude Question SOLUTION: 3 numbers are to be choosed from 1 to 9 in such a way there always becomes an A.P series. Find the probability?

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3 numbers are to be choosed from 1 to 9 in such a way there always becomes an A.P series. Find the probability?

Asked In TCS TargetTCS 2014 (10 years ago)

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TargetTCS 2014  (10 years ago)

for first question total cases 9C3=84.. 123,234,345,456,567,678,789,135,357,579,246,468,147,258,369,159, total 16*2 because reverse is also possible so 32/84..correct if i am wrong...

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Vishnu Vijayan  (10 years ago)

THE POSSIBILITY FOR AN AP SERIES ARE,
123,234,345,456,567,678,789.....(1 DIFFERENCE)
135,246,357,468,579......(2 DIFF)
147,258,369.......(3 DIFF)
159.......(4 DIFF)
SO TOTAL NUMBERS THAT CAN BE FORMED=
TOTAL PROBABILITY=9C3=84
SO THE ANSWER IS 16/84=4/21

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Hem Chandra Joshi  (10 years ago)

we all know that difference between two numbers in A.P is same.
we have to select digits from 1 to 9.
so total cases=9C3=12*7.
favourable cases: when difference is 1 =7,when difference is 2 =3, when difference is 3 =1 and when it is 4 =1.
total favourable cases=12.
so probability=12/(12*7)=1/7.

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Riya Khanna  (10 years ago)

the possibility for an AP series are,
123,234,345,456,567,678,789( with difference 1)=16(reverse is also possible)
135,246,357,468,579( with difference 2)=10(reverse also)
147,258,369( diff 3)=6(reverse also)
159 (diff 4)=2 (reverse also)
the sum will be 34
and total probabilty= 9C3=84
so, prob= 34/84

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pankaj kandpal  (10 years ago)

now for sample space it would be 9C3
and now count ap's having d=0, i.e. 1,1,1 2,2,2 ... 9,9,9
so total 9 ap's with d=1 we can have 123,234,345...
total of 6...now ap's with d=2 135,246... toatl of 5...now for d=3
it would be...3...for d=4 it would be only 1..so tatal of 15+9=24
now except all ap's with d=0...oter can be reversed...so total=30+9=39
so probability=39/(9c3)=39/84;

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pankaj kandpal  (10 years ago)

why not including ap's with d=0...i.e. 1,1,1 2,2,2 etc

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hitesh kapila  (10 years ago)

why not d=0 taken

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