Choosing the same number is (1/4)^4 since say our chosen # is 2 then prob. of first person choosing 2 is 1/4 etc. Now here we want different numbers so let first person choose 1 with prob. (1/4), second chooses number 2 with p = 1/4 so I think that probability is the same with the first case and p=(...

Q2) My ans. is 17 Look at 3 first --3-- we have 5 cases 12345, 12347, 12356, 12357, 12367 -3--- first place could be 1 or 2 while last 3 places could be 456 or 457 or 567 so 2x3= 6 cases 3---- only with 4567 1case now take 3 out of the picture and look at 4 --4-- last 2 digits could be 56 or 57 so 2...

1) x-3 = 4k => x = 4k + 3 put k=3,4,5,6, etc. and get 15,19, 23 etc so remainders (when divided by 12) are 3, 7, 11 (different so A is insuff.)
2) x-27 = 36k => x= 36k+27 = 3*12k+27 = 3*12k+3*12+3 =
=3m+3 so remainder is 3 when x is divided by 12 (same so B sufficient)

be careful here:
sqrt(4) = 2
sqrt(9) = 3
sqrt(4) + sqrt(9) = 5
but sqrt(4+9) = sqrt(13) <>5
put some values to check
thanx

first of all I think you mean x^2 = y^2 (1). Now since we have squares
we have 2 cases for example (-3)^2 = (3)^2 and equation (1) holds
for y=+,- x and not just for positive x only, so ratio could be either 1 or -1

prob. of working (1-x) so prob of N working is (1-x)^N and set will not work when at least one fails and --> p=1-(1-x)^N

f(a)=-3a, f(b) = -3b, f(a+b) = -3(a+b) = (-3a)+(-3b) = f(a)+f(b)