1is the following answer correct p2(1-p)k-2
Q:=> Two numbers a,b are chosen from the set \left\{2,2^{2},......,2^{25} \right\}. Find the probability that log_{a}b is an INTEGER.
Q:=> An instrument is being tested, upon each trial the instrument fails with probability p. After the first failure the instrument is repaired and after the second failure it is considered to be unfit for operation. Find the probability that the instrument is rejected exactly in the kth trial.
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7 Answers
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! :)
1first 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
