62See the most important thing si that
ΣP(Ek) = 1
hence (n+1).c=1
Thus, P(Ek) = 1/(n+1)
now we have P(0)=1/(n+1)
P(k) = (k+1)/n+1
P(k) is the probability that atmost k students have failed...
21Thnx a lot SirJee and RICHA!! [1]
21
tapanmast Vora
·2009-04-09 11:24:40
OH got it now!!! finally [2] [1] [1]
Richa edit this un : p(a)=Σ1/n+1(k/n) it shud be p(a)=Σ[(1/(n+1))*(k/n)]
1see p(a)=p(a/ek)p(ek) rite....
now p(ek)=c=1/n+1 kkk
n now p(a/ek)=k/n
lyk if it s given 2 students pass
it means p(tht student choosen is passed is 2/n na
so p(passing /k students pssed is k/n)
now summation k=o to n u ll get the ans
1
vector
·2009-04-09 11:16:37
i m sorry fr mistyping edited
21
tapanmast Vora
·2009-04-09 11:15:38
i m stioll confused [2]
can u give the logical statement for ur P(A)
11yeh TMH ka problem hai
A VERY BAD 1 !!!!!!!!!!!!!!!!
i completely ununderstood that
1
vector
·2009-04-09 11:10:43
main bhi yahi bol rahi thi[1]
1
vector
·2009-04-09 11:08:28
after c value it slyk
1/n*1/n+1+2/n*1/n+1........n/n*1/n+1
1
vector
·2009-04-09 11:05:29
n as p(ek)=c
c+c+......c(n+1 times)=(n+1)c=1
after tht i think its easy
21
tapanmast Vora
·2009-04-09 11:05:27
hmmm value of C i too got,
Sir is solving it , wait 4 sumtime
1
vector
·2009-04-09 11:03:35
i gotthoda thoda lyk the events r exhaustive n mutually exclusive
so p(e0)+p(e1)+...............p(en)=1
21
tapanmast Vora
·2009-04-09 11:02:58
Baye's has been BOOING me off-late [2]
11nish. bhaiyya......pls. help !!!
11
rkrish
·2009-04-09 10:53:58
me too...even i didnt understand [2][2]
21
tapanmast Vora
·2009-04-09 10:52:53
Can sum1 give a mor clarified version, i din eXACTLY get wat gordo said in da abv link
11
rkrish
·2009-04-09 10:43:39
http://targetiit.com/iit_jee_forum/posts/mathematics_4452.html
post #13
1
vector
·2009-04-09 10:39:51
dis ve been posted in tiit by manipal
