1considerin a and b to be distinct,
a and b will be of the form 2^m and 2^n then for the log function to be an integer n/m must be integer where n,m belong to {1,2,...,25}. so possible pairs of (m,n)= (1,2), ....,(1,25) ---> 24 pairs
again (2.4),....,(2,24) --> 11 pairs
again (3,6).......(3.24) --->7pairs
again (4,8).......(4,24) --->5 pairs
again (5,10)....(5,25) ----> 4 pairs
(6,12)....(6,24)..... 3pairs
(7,14), (7,21)..... 2pairs
with 8 ----- 2 pairs
with 9 --- 1 pairs
with 10 --- 1 pair
with 11 --- 1pair
with 12 -- 1 pair
so reqd probability= 24+11+...+1/ 25C2= 31/150
492) the instrument fails 2nd time at kth attempt to be rejected..!
it wont fail if it has not failed once in the previous k-1 attempts.
and it wont fail if it fails for 1 time in k-1 attempts but does not fail for the kth time! :)
this was a hint..i had solved it in this way!!
gave the answer!
so...try it out! :)
1
thumbsdown007
·2011-02-21 23:40:10
is the following answer correct p2(1-p)k-2
23(k-1)p2(1-p)k-2 ??????????????
1
bindaas
·2011-02-26 01:32:32
first one :
\log_ba=\frac{\log_2a}{\log_2b} \\ \texttt{So the problem reduces to finding the probability of }\frac{a}{b}\\\texttt{being an Integer}:\left\{a,b \right\}\in\left\{1,2,3,\cdots25 \right\}\\ P=\frac{62}{\binom{25}{2}}=\boxed{\frac{31}{150}}
EDIT :OOPS POST 2 IS SAME AS THIS
1
bindaas
·2011-02-26 13:33:59
My first Latex Document :P
