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

• 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...
• 11 years agoHelpfull: Yes(45) No(6)
• 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
• 11 years agoHelpfull: Yes(8) No(10)
• 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.
• 11 years agoHelpfull: Yes(1) No(1)
• 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
  
• 11 years agoHelpfull: Yes(1) No(2)
• 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;
• 11 years agoHelpfull: Yes(0) No(13)
• why not including ap's with d=0...i.e. 1,1,1 2,2,2 etc
• 11 years agoHelpfull: Yes(0) No(7)
• why not d=0 taken
• 11 years agoHelpfull: Yes(0) No(2)