Dear Varsha ,
2nd ques is quite straight forward as
we can send  the letters and envelopes in 4! different ways out of which there will be only one way that letters are mailed properly wid der envelopes then
according to d quest, all letters are not mailed properly which means total ways minus one way of proper mail,
p = (4! - 1)/ 4! = 23/24
ans = (c)
 
						

							
Dear Varsha ,
Q 1
if d coin is tossed n times ,den total possibilities will be 2^n.
let chance of getting r head out of n be nCr
den total way of getting odd times head will be nC1 + nC3+ nC5 +.......nCn (if n is odd) 0r nC1 + nC3+ nC5 +.......nC(n-1) ,for n is even
nw, (1 + x)^n = nC0+ nC1 x + nC2 x^2+........
put x=1 , 2^n = 1 + nC1 + nC2 +......
put x = -1, 0 = 1 - nC1 + nC2 - nC3 +......
just take difference of d above 2 eqn ,
2^n = 2( nC1 + nC3 +........)
if n is odd den
nC1 + nC3+ nC5 +.......nCn = 2^(n-1), probability,p = 2^(n-1)/2^n = 1/2.
 
						

							
hi varsha ,
Q2
toatal no of ways in which 4 letters n  4 envelopes can be arranged is 4!
der can be only 1 way in which in which all letters are mailed properly wid der envelopes.
so required probability,p = (4!-1)/4!   [ all d letters not mailed properly = total ways - no of way in which dey r mailed properly ]
p = 23/24