special question of part A

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Q-In a bag containing only blue , red & green marbels all but 15 are blue , all but 13 are red & all but 12 are green . How many are red ?
(a) 13
(b) 7
(c) 25
(d) 20
Q-A coin is tossed 6 times . the probability that heads will occur at least once is ?
(1) 63/64
(2) 1/3
(3) 1/64
(4) 3/2

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Q-In a bag containing only blue , red & green marbels all but 15

allk but means not
all but 15 are blue means 15 are not blue .so it may be red+green=15
blue+green=13
blue+red=12
r+g=15
b+g=13
----------
r-b=2
r+b=12 .so 2R=14 r=7.like wise calculate remaining 

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how that r-b came? pls

how that r-b came? pls explain that and also post some more tricks to solve this type of questions if you know ...

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plz post some more questions

plz post some more questions apart from the model questions given by csir. thnx for starting dis forum..cud b quite useful

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ans

Q-In a bag containing only blue , red & green marbels all but 15 are blue , all but 13 are red & all but 12 are green . How many are red ?
(a) 13
(b) 7
(c) 25
(d) 20
Q-A coin is tossed 6 times . the probability that heads will occur at least once is ?
(1) 63/64
(2) 1/3
(3) 1/64
(4) 3/2

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answers

(a), (3). Thanks

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chaudhury

i think ur ans is correct . can u explain plz how these equations are solved of ball questions. 

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sorry for late

sorry friends . the correct ans of these questions actualy i don't know .these questions are asked in sampel paper of new revised syllabus of part A downloaded by csir . 

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pls post ans too thank u

pls post ans too
thank u

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1 & 3

1 & 3

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 -A coin is tossed 6

 -A coin is tossed 6 times
1st time (H.T) so probability of head 1/2
2nd time (H,T) so probability of head 1/2
3rd time....................................................1/2
4th time.....................................................1/2
5th time.......................................................1/2
6th time .......................................................1/2
so probability of head atleast once ...1/2*1/2*1/2*1/2*1/2*1/2=1/64
ans 3