DSE 2011 A (a request to duck/Amit Sir, Q40 plz)

[Sinistral]

Ques no 35



what is wrong in the following solution as the correct ans is .(b) equal to the probability that it will end in 6 games.

[Sinistral]

[Sinistral]

[laracroft]

Hi Sinistral

for the Q. 8
if u carefully calculate, u ll get solutions as = 10 (1/2)^5 for both, 6 games as well as 7 games.. :)
hence option B

[Sinistral]

i think i made 1 mistake in my earlier solun.

is this solution correct???

f(6)= 2* [5C3]*(1/2)^5   * (1/2)
f(7)= 2* [6C3]*(1/2)^6   * (1/2)

but still f(6) not equal to f(7)

[laracroft]

Hi sinistral...

u got it right this time... but u get the answer same for 6 and 7 matches... :)
plz chek yr calculations once more ..!


Plz let me know if u figure out the rest of the two questions that u posted !

[Sinistral]

phew!!! finally f(6)=f(7) :)

and rest of the 2 ques please...

[Sinistral]

duck, wolverine or any one else,
could u please help me with ques no. 36 & 40

[Sumit]

Q36. b).747
Expalanation: put 1 green ball & zero red ball in first box and 49 green balls & 50 red balls in 2nd box..

So, Prob.=1/2*(1)+1/2*49/50=0.747.

[Ayushya Kaul]

Bump. Had a problem in the same aforementioned questions!

[Ayushya Kaul]

How did you get the probability distribution function?

[Ayushya Kaul]

Could you explain how you got the answer for question 8?

[duck]

Q36) Lets, suppose he puts (50 R) in Bag 1 and (50 G) in bag 2.
So, probability of A being admitted = 1/2

But, we've a better way:
Put (50R + 1G) in Bag 1 and (49G) in Bag 2.
Probability = (1/2)(1/51) + (1/2)

But again, we've a better way:
Put (50R + 2G) in Bag 1 and (48G) in Bag 2.
Probability = (1/2)(2/52) + (1/2)
........
........
and so on...

Thereofore , the Best way:
Put (50R + 49G) in Bag 1 and (1 G) in Bag 2
Probability = (1/2)(49/99) + (1/2)

So, the idea is we can always ensure a Probability of (1/2) by keeping only the minimum number of green balls in Bag 2. And hence, can maximise the probability in the above way.

[Sinistral]

thank u so much duck and Sumit 

now after looking at the explanation, the distribution seems fairly logical. :) thanx once again.

plz help me with student and the no. of tests he can take prob as well, Q40.

[Ayushya Kaul]

How did you arrive at these calculations?

[Sinistral]

first of all the solution to ques no 8 given in the very first post is wrong (as it is clear by reading the subsequent replies)

the right ans is in the 4th reply

to calculate how the game finishes in exactly 6 or exactly 7 matches, keep in mind that the game has to be stretched till the last match ==> the concerned player SHOULD win the LAST match AND he should have already won EXACTLY 3 of his previous matches (in this case 3 of his previous 5 or 6 matches)

for example calculating for x=6
5C3(1/2)^5: choosing 3 out of his previous 5 matches and winning those (& obviously losing the remaining)
1/2: probability of winning the last match.
2 : this can be done for both the players.

multiplying the above gives:
f(6)= 2* [5C3]*(1/2)^5   * (1/2)

similarly for f(7)

note that f(6)=f(7)

[Ayushya Kaul]

Thanks mate. Cleared that up well. Cheers!

[Erika]

very well explained. thumbs up

[madhur1987]

hey sinistral whats the answer for que 40

[Sinistral]

ans is b

Explanation by Amit Sir

A more detailed explanation