DSE 2011 QUES NO. 36

[sunny2009]

in order to join the "gamers club',mr. A  must choose a box from the 2 identical boxes in a room, and draw 1 ball from the chosen box.all he knows is that both boxes r nonempty,and have a mix of red and green balls.if the ball that he draws from his chosen box is green,he is admitted to d club.U r given 2 identical boxes,50 red balls and 50 green balls,and asked to allocate these balls to d 2 boxes in order to maximise mr. A's probability of being admitted to d club,given that he will choose a box randomly.IF u allocate these balls correctly,d probability that mr. A gets admitted to d club equals whatttt???????????????

[komal]

in first box put 1 red ball and 0 green balls, in second 49 red balls and 50 green balls, prob will come out to be .747( 1/2 *1 + 1/2 (49/99))

[sunny2009]

@komal
in calculating probability,why dont we take into account d green balls being drawn???
i didnt get ur answer..will u plzz explain how we get the answer in some more steps..

[sonudelhi]

i did it as.......let first box x red balls and y green balls.then second box would have 50-x and 50-y respectively.

P(G)=P(b1)P(G/b1)+P(b2)p(G/b2)
      =1/2*y/x+y + 1/2*50-y/100-x-y

now maximize P(G) subject to conrs. x+y=1

you will get y=1 and x=0 implies that 0 red and 1 green ball in first box

[komal]

oops, i mistakenly took red ball selection for entrance, if selection of green ball is necessary for entry into the club then allocation should be 0 red ball and 1 green ball and rest in other box, so prob of selection of green ball = prob box I is selection * prob of green ball being selected given first box is chosen + prob second box is selected * prob of  green ball is selected when second box is selected = 1/2* 1 /1  +  1/2 * 49/99= .747

[sunny2009]

thank u so much komal and sonu...!!
thanksssss a ton :)

[Chinni18]

Thanks Komal and Sonudelhi. This was a point of great confusion for me

[aditi5000]

Hi Komal, but the qustion says "both boxes non empty and have a mix of red and green"
so are you sure we can allocate 1g and 0r in any box? in this case answer comes to 0.5 :(

[Chinni18]

Aditi, even if there is one ball in the box, it satisfies the "non-empty" restriction.
So the probability is:
P(Chossing 1st Box)*P(Choosing green ball|B1 is chosen) + P(Choosing 2nd box)*P(Choosing green ball|B2 is chosen)
=(0.5)(1) + (0.5)(49/99)
=0.747

[Chinni18]

Hmm but you're right, in this case the mix of red and green balls clause isn't satisfied at all. So 0.747 cannot be the answer because we're not allowed to not mix the balls.
Argh, there goes my conviction

[sunny2009]

@chinni and aditi
what i thought was this....
"the condition that both d boxes r non empty and have a mix of both green and red balls is the initial condition".....and now "we r to leave this criteria and then focus solely on "allocating balls such that we maximise d probability of mr. A being admitted to d club"...n "we can do this only by putting balls in d way described above.......so the answer is correct in my view....why are we going on initial condition?????
..correct me if i m wrong...........thanks!!

[Mr. Nobody]

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[aditi5000]

yeah i'm also stuck there, how are they equal...
sunny@ yeah i guess that makes sense... that both boxes contain a mix is what is told to mr a, not to us... we are told to max probability of drawing a green ball... makes sense, sort of