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The no.s (a,b,c) are selected by throwing a dice thrice.then the probability that (a,b,c) are in A.P. is?
i am geting the ans to be 18/216 but the ans is 21/216
Hi Debadutta,
(a,b,c) can be
(1,2,3) -------- 2 possibilities (3,2,1) is the other order
(2,3,4) -------- 2 possibilities
..... --------- -do-
.....
(4,5,6) -------- -do-
it could be
(1,3,5) -------- 2 possibilities
(2,4,6) -------- 2 possibilities
or it could be
(1,1,1)
(2,2,2)
....
(6,6,6).
So total number of ways = 2*6 + 6 = 18.
Your answer is very fine.
Have confidence in your working. It helps for competitive exam prearation.
21/216 is not the right answer.
Regards,
Ashwin (IIT Madras).
How can This be an A.P.Minimum C.D for an AP is 1.
(1,1,1) (2,2,2)........(6,6,6).
the answar is 18/216, there cant be more than 18 combos of AP for 3 die
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