duck: please see this.

[Mauli]

http://discussion-forum.2150183.n2.nabble.com/2006-dse-q31-tp6643980.html;cid=1371101202334-835

the contents to this thread have been deleted.
please help me with q31-33. i am stuck badly.

[Abhyudaya]

Hi Maullii .... m getting 7/16 as d answer.... sry

[MR]

Hey Mauli please tell me which chapters are to done from the book by kapoor and gupta for probability distributions? I arranged the book but there are so many chapters on distributions. :(

[Mauli]

i think  till the chapter on expectations will be sufficient.
however, that is my personal belief.
i think that covers all.
the rest like sampling and all they have already been given in the basic book of sc gupta in a very nice way.
i did this!so , ya i feel tht is more than enough.

[Mauli]

how?
can u  please share your method.
maybe some small mistake and we can discuss and sort it out.

[MR]

Thanks Mauli. :)

[duck]

Hi Mauli.. :)

Q31) P(A winning) = B wins the first match and then A wins in the subsequent matches
                           + A wins in the first match and then A wins in the subsequent matches.
                         = [BCAA+BCABCAA+BCABCABCAA+....]
                           +[AA+ACBAA+ACBACBAA+.........]
                         = [(1/2)2 + (1/2)5 + (1/2)8 + (1/2)11 +…]
                           +[(1/2)4+ (1/2)7 + (1/2)10 + ……]
                         = 4/14 + 1/14
                         = 5/14

Q32) As A and B are symmetric players . Probability of winning must be the same.
P( B winning) = 5/14.
Therefore, P(C winning) = 1- P(A winning) - P(B winning)
                                 = 1-5/14-5/14
                                 = 2/7


Q33) If the game continues indefinitely ie.
like ACBACBACB....
then, its same as finding limit (as n tends to infinity) (1/2)^n, which is equal to zero.


 

         

[Mauli]

cool:)!
thankyou duck.
thanks a lot!