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Question: In a game of chance, first 20 natural numbers are listed on a piece of paper and three numbers are chosen at random. If the numbers form an A.P. you will get Rs. 10000, If the numbers form a GP you will get Rs. 20000, otherwise you have to pay Rs. 8000. What is the mean expected gain and the corresponding risk, if you play the game?–Free Chegg Question Answer

In a game of chance, first 20 natural numbers are listed on a piece of paper and three numbers are chosen at random. If the numbers form an A.P. you will get Rs. 10000, If the numbers form a GP you will get Rs. 20000, otherwise you have to pay Rs. 8000. What is the mean expected gain and the corresponding risk, if you play the game?

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ways of choosing 3 numbers out of 20 is C(20,3)=1140  [(a,b,c) with 1 <= a < b < c <= 20]

finding the series of GP,

Sets with common ratio 2/1: (1,2,4) , (2,4,8) , (3,6,12) , (4,8,16) , (5,10,20)

Sets with common ratio 3/1: (1,3,9) , (2,6,18)
Set with common ratio 4/1: (1,4,16)
Sets with common ratio 3/2: (4,6,9) , (8,12,18)
Set with common ratio 4/3: (9,12,16)

# of such sets = 11

Probability of GP= 11/1140

finding series of AP,

with common difference 1: (1,2,3).(2,3,4)…………(18,19,20) so, # terms=18

with common difference 2:(1,3,5),(2,4,6)………….(16,18,20 so, # terms=16

with common difference 3:(1,4,7),(2,5,8)…………(14,17,20) so,#terms=14

with common difference 4 :(1,5,9),(2,6,10)……….(12,16,20) so # terms=12

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similarly this way,

last will be

with common difference 9:(1,10,19),(2,11,20) so, # terms=2

total terms of AP=18+16+14+12+10+8+6+4+2=90 terms

probability of three numbers being in AP=90/1140

probability of losing=1-90/1140-11/1140=1039/1140

AP(win) GP(win) lose probability gain/loss 90/1140 11/1140 039/1140 -8000 20000 mean expected gain-(90/1140) 10000+(11/1140)*

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