21COZ if the relation x<y doesnot exist then our ans wud easily double up and come down to "8" wich is a part of the options
Here i will be posting ques on probability which could prove to be useful 4 every1
Do try to give explanations if u can
options will be given on demand[1]
Q1
If the papers of 4 students can be checked by any one of 7 teachers , then the probability that all the 4 papers r checked by exactly 2 teachers
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UP 0 DOWN 0 0 80
80 Answers
11@tapan post #60
but u said answer is A ie 5 na ..................
11Q4
A box contains N coins,of which m r fair and the rest r biased .The probability of getting head when a fair coin is tossed is 1/2 ,while it is 2/3 when a biased coin is tossed .A coin is drawn from the box it shows tail .The probability that the coin drawn is fair is........
2111yup the ques is perfect
mere ko sir ne orally bataya bhi tha
par main bhool gaya[2]
21dhundhni padegi........
but thats the closest.......
pl. chk all ur specifications abt x,y are crrct in regard to x<y especially
21x = 2007y/(y-2007)
now let y be 2007 + p coz x cant b negative
then x = 2007 + 20072/p
now chk diff values
11yaar i dont have the answer to this 1
please post ur explanation!!!!!!!!!!!!!!!!
21second :
y = 2007*10
x = 2007*10/9
3rd :
y = 2007*2008
x = 2008
4th :
y = 2007*224
x = 2007*224/223
21tell me agar sahi aa rahe hai!
to i'll post more values
First :
y = 2007*4
x = 2007*4/3
21MANI : [16] [16]
I TOLD Naa BHAI for Q1 it had to b wat I said ther was no oder Option [16]
neways chalta hai no probs [1] [4]
11i appreciate ur work MR.MOHD [1]
NOW
Q3
The positive integer pairs (x,y)such that 1/x+1/y=1/2007,x<y,is???
1no yaar ur absolutely correct...........i have posted my answer above but u gave the solution nice work........gud job.....(i dont knw ur name)...........
1for d product to b divisible by 3, atleast one of d two no.s must b divisible by 3...
>> let first no. b 3 .... second one can b any of d remaining 99 no.s.... hence total no. of selections possible = 99
>> let first no. b 6 .... second one can b any of d remaining 99 no.s... but since (3,6) is already considered in first case, hence no. of selections here = 98
>> first no. = 9 .... total selections possible = 97
.
.
.
.
>> first no. = 99 .... total selections possible = 67
hence probability = 67+68+...+99 / 100c2
= 83/150...
most informal method, but still... easy approach...!!![3]
let me know if i'm wrong anywhr...!!!
1hey manipal is the answer C)------>83/150.........plzz correct me if im wrong.........
11ABHI BHAI
PERFECT ANSWER
PLEASE GIVE THE PROCEDURE TO SOLVE THIS PROBLEM
I AM GLAD THAT AT LEAST U TRIED[1]
11Comprehension - 1
A class consists of n students. For 0 ≤ k ≤ n, let Ek denote the event that exactly k student out of n pass in examination. Let P(Ek) = pk and let A denote the event that a student X selected at random pass in the examination.
11. If P(Ek) = C for 0 ≤ k ≤ n, then P(A) equals
(A) 1/2 (B) 2/3 (C) 1/6 (D) 1/(n+1)
12. If P(Ek) = C for 0 ≤ k ≤ n, then the probability that X is the only student to pass the examination is
(A) 3/4n (B) 2/(n+1)
(C) 2/n(n+1) (D) 3/n(n+1)
13. If P(Ek) k for 0 ≤ k ≤ n, then P(A) equals
(A)3n/(4n+1) (B)3n+1/3n
(C) 1/n+1 (D) 1/n2
14. If P(Ek) k for 0 ≤ k ≤ n, then the probability that X is the only student to pass the examination is
(A)3/n(n+1) (B) 6/n(n+1)(2n+1)
(C)1/n (D) 1/n(2n+1)
11PERFECT BHAI
MEKO PROBABILITY MEIN BAHUT TARAKI HAI
CHAL BAAD MEIN AUR QUES POST KAROONGA
13after picking up one ball
remaining balls of favourable choice
n-1 of same colour
2 balls of other colour but the same no.
balls remaining 3n-1
req. probab
(n-1+2)/3n-1
13http://targetiit.com/iit_jee_forum/posts/li_l_challenge_5_3951.html
check dis one...simillar Q
11
x1+x2+...+xk=n - k/2*(k+1)
M=n-k/2*(k+1)+k-1=n-k/2*(k-1)+k-1
bhai yeh thoda bhari ho raha hai
please explain kar do ki
1st step mein -k/2 X (k+1) kyun kia hai
aur 2nd step mein kya kia hai
sorry 4 this silly ques
i know i am a spoil sport
13MCk-1 is the no. of integral solutions of
x1+x2+...+xk=n - k/2*(k+1)
M=n-k/2*(k+1)+k-1=n-k/2*(k-1)+k-1