!!Please comment on the correctness of this question!!
In a lucky draw competition, a contestant had to draw 3 tickets from a pile of well-shuffled tickets. If the numbers on these tickets were in arithmetic progression, he would be adjudged the winner of the contest. If the numbers were not in arithmetic progression, he would return them to the pile. The tickets were numbered from 1 to 201. What is the minimum number of contestants to ensure that at least one of them wins the contest? If 100 people took part in the lucky draw, do you think we could take for granted that at-least one contestant would win the prize?
10 comments:
And one more thing,, if one person takes his tickets, we would say that the # of remaining tickets is 198 or 201. (If its 198, then 100 people cannot participate and if its 201 then we can sat that there is no certainty that a person will be able to pick an AP)
One review of the question(if the question is absolutely correct)
Its too tough and requires detailed explanation.
If a person picks up 3 chits, and if they are not in A.P., he returns them back to the pile.
ananth if the chits are returned back to the pile then each condition is same as the first one which would be u never know. it can be an ap or nt . and also the moment u get an ap, winner is declared so i guess play stops.....
however my basic doubt is that
is there any other constraint which wud increase the possibility of an ap with increased no. of contestants? maybe the answer to this question holds the key to solving this puzzle, but i dnt get it.
here are some constraints i have come up with which may explain the question but then i dnt know if this is a part of the question:
1) After some1s chance is over he tells wht all nos he got to the second person(in case his is nt an ap)
2)The second person can make out a difference between a a used chit and an unused chit...
i mean im just guessing and this is also what kind of constraints i was talking abt.
So please review the question ananth.
And 1 more thing if there is no information like this required then i give up.
My logic behind solving the question was finding the probability that the 3 numbers taken out randomly were in A.P. and then finding 1/probability.
ananth please post the solution
The tickets were numbered 1 to 201.
Let's suppose the participant picks up one of the chits as 1, then the possibility of an A.P. is eliminated.
If the contestant picks one of the chits as 2, then there may be an A.P. as (1,2,3).
If the contestant picks up one of the chits as 3, then there are two favourable cases: (1,3,5) and (2,3,4).
Thus the number of favourable instances increases as 0, 1, 2, 3.. and so on. One can also understand that these will increase till 99 (one less than the middle number) and will then decrease back to 0 when the last number, 201 will be selected.
Therefore, the number of favourable outcomes is 0 + 1 + 2 + 3 + .... + 98 + 99 + 98 + 97 + .... + 2 + 1 = 9,801.
Also, the total number of outcomes is C(201, 3) (where C(n,r) denotes the number of possible combinations of n objects taken 3 at a time) = 13,33,300.
Hence, the probability that the numbers are in A.P. is 9,801/13,33,300 = 0.00735.
The idea of finding the minimum number of people needed to definitely pick up an A.P. can be used only if one considers that the numbers picked up form a different triplet each time. Sorry for overlooking this fact.
If this is taken into account, the n minimum number of people comes out to be 136.
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