@maddy2807 said:9?
I am confused with the Ans.. I read somewhere that the OA:7 ..But i Think it should be 9..Since many other well-known-quant-killers also came to this answer
I second you..Please explain your approach then i will share mine
Approach please....a miserable person when it comes to probability here 
@ScareCrow28 said:I am confused with the Ans.. I read somewhere that the OA:7 ..But i Think it should be 9..I second you..Please explain your approach then i will share mine
took all the cases...
(9C3+9C4+9C5+9C6+9C7+9C8+9C9)/512
in this case prob will be more than .9
but one confusion i hv is. if target is destroyed in 3 shots. how can we hit it more than 3times.
@maddy2807 said:took all the cases...(9C3+9C4+9C5+9C6+9C7+9C8+9C9)/512in this case prob will be more than .9but one confusion i hv is. if target is destroyed in 3 shots. how can we hit it more than 3times.
@ScareCrow28 said:I am confused with the Ans.. I read somewhere that the OA:7 ..But i Think it should be 9..Since many other well-known-quant-killers also came to this answerI second you..Please explain your approach then i will share mine
http://www.pagalguy.com/forums/quantitative-ability-and-di/official-quant-thread-c-t-44541/p-1674757/r-1722099?page=374
Check out naga sir's solution and the other solutions as well..
I could not figure it out..maybe you guys can
Oh...my bad my bad 😛
@ayushnasa said:If we do it the other way round with 7..that is find probabilty of it not destroying the target...it comes out to be (7c1+7c1+7c2)/512=29/512=0.05.. So the probabiltty of destruction is nearly 0.94.. for 7 throws...@ScareCrow28
It shoulod be(7C0+7C1+7C2)/2^7 na? why 512??
@soumitrabengeri said:www.pagalguy.com/forums/quanti...Check out naga sir's solution and the other solutions as well..I could not figure it out..maybe you guys can
yeah kya soln hai... sab upar se gaya...
@ScareCrow28 said:It shoulod be(7C0+7C1+7C2)/2^7 na? why 512??
Deleted the post already....I have gone insane...need a break
@maddy2807 said:yeah kya soln hai... sab upar se gaya...
Even i could not figure it out..that's why i asked you guys to take a look..udhar OA has been mentioned as 7
@ayushnasa said:Deleted the post already....I have gone insane...need a break
By doing it your way.. We have figured out that in 7 attempts..we are not getting the answer..but if we put n=9
Then (9C0+9C1+9C2)/512 = 0.08
This is satisfying.. I dnt know howcome naga sir got 7
But one thing is still undoubtedly doubt-able here.. Why would we hit it 4 times if we have destroyed it in 3 times already?? 
@ScareCrow28 said:By doing it your way.. We have figured out that in 7 attempts..we are not getting the answer..but if we put n=9Then (9C0+9C1+9C2)/512 = 0.08This is satisfying.. I dnt know howcome naga sir got 7
Even GODs make mistakes 
Maybe he did :mg:
Maybe he did :mg:@ScareCrow28 said:Probability that I hit a target is 1/2. The target is destroyed if 3 shots hit the target. What is the minimum number of shots that I should try, so that the probability that I destroy the target is more than 0.9 ?1) 62) 73) 84) 9
Why dont we ask him directly?? @naga25french saar, could you explain this question? You had got OA as 7. Everyone here seems to get 9...
In the meantime:
We have 3 pastures with grass of identical height, density and growth rate. The first is 3 1/3 acres and can feed 12 oxen for 4 weeks. the second is 10 acres and can feed 21 oxen for 9 weeks. The third is 24 acres. How many oxen can be fed on these pastures for 18 weeks?
@ScareCrow28 @soumitrabengeri @ayushnasa
if I can deduce well
acc to naga sir...
since the prob to hit the target is 1/2
so if we hit 6 target. obviously 3 will hit the target. and 1 more is taken to ensure that the prob is achieved. otherwise it wud hv been 1.
but still with this method. prob remains less than .9
please suggest frnds.
@ScareCrow28 said:In the meantime:We have 3 pastures with grass of identical height, density and growth rate. The first is 3 1/3 acres and can feed 12 oxen for 4 weeks. the second is 10 acres and can feed 21 oxen for 9 weeks. The third is 24 acres. How many oxen can be fed on these pastures for 18 weeks?
options?
@maddy2807 said:@ScareCrow28@soumitrabengeri@ayushnasaif I can deduce wellacc to naga sir... since the prob to hit the target is 1/2so if we hit 6 target. obviously 3 will hit the target. and 1 more is taken to ensure that the prob is achieved. otherwise it wud hv been 1.but still with this method. prob remains less than .9please suggest frnds.
(Can't figure out "+1") You said its obviously destroyed in 6 attempts then why "+1" Ab isko @naga25french Sir he bataye to achha hai :P@maddy2807 said:@ScareCrow28@soumitrabengeri@ayushnasaif I can deduce wellacc to naga sir... since the prob to hit the target is 1/2so if we hit 6 target. obviously 3 will hit the target. and 1 more is taken to ensure that the prob is achieved. otherwise it wud hv been 1.but still with this method. prob remains less than .9please suggest frnds.
This is exactly the same thought process i went through when i read the post..confused..
@ScareCrow28 said:In the meantime:We have 3 pastures with grass of identical height, density and growth rate. The first is 3 1/3 acres and can feed 12 oxen for 4 weeks. the second is 10 acres and can feed 21 oxen for 9 weeks. The third is 24 acres. How many oxen can be fed on these pastures for 18 weeks?
18?