# This probability problem is killing me HELP!!!

A bag contains a coin of value "M" and a number of other identical coins whose aggregate value is 'm' .A person draws 1 coin at a time(WITHOUT REPLACEMENT) till he gets 'M'.find the value of his expectation??????????

there are total 'n' coins......

plssss do it thanking in advance for this......

Note by Riya Gupta
5 years, 9 months ago

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If he get M the first time then expectation value is M(1/n){The value * probability} If he gets in 2nd try(i.e. 1 m coin and one M coin) then expectation value is (M+(m/n-1))((n-1)/n)*(1/(n-1)) Try writing this series for 3 tries and so on. You will get a series. Sum it and u will get the answer. srry cant use symbols. New to this site and not comfortable using symbols.

- 5 years, 9 months ago

okkk thanks so much

regarding new to ths site i can undrstand ur problm i too joind brilliant ds week only :)

no prolm

u can use formatting guide may b it'll hlp you :)

- 5 years, 9 months ago

is no one there to clarify my doubt? plzzz HELP!!!

- 5 years, 9 months ago

is the answer M+(mn/2) ? Here expectation value is of the money picked out right?

- 5 years, 9 months ago

plssss can u xplain it :/

- 5 years, 9 months ago

correct answer is M+(m/2) jz chckd it rite nw

- 5 years, 9 months ago

There is no n factor? should be there as it is given. is n=1 then expectation is M. Otherwise it will be different.

- 5 years, 9 months ago

The answer u gave is correct. had missed a (1/n) factor .

- 5 years, 9 months ago