rkrissh ...ur "drona of mathematics" yaar.......
Out of 46 consecutive integers two are chosen...find the probability that their sum is odd
-
UP 0 DOWN 0 0 18
18 Answers
rkrish...now plz ans this one....more simplified...
Out of 2 consecutive integers two are chosen...find the probability that their sum is odd
actually we can not pick 1 twice out of 1,2
thatz the mistake i did
thanx rkrish...
i think i got it...we can not pick the same no. twice...as we r taking it simultenously.....not one after another..
arrey yaar.....its not like that...dekho....
why 1st one & 2nd one...
even & odd ya odd & even kya fark padta hai.
you have to choose two nos.
one odd no. out of 23 & one even no. out of 23.
kitne hue ?? 23C1 for odd & 23C1 for even.
So, fav case hua 23C1 * 23C1 = 23 * 23
Abhi sample space toh nahi badlega....46 nos. mein se you have to choose 2 nos.
So sample space = 46C2 = (46*45)/2 = 23*45
Hence P = (23*23)/(23*45) = 23/45
picking up numbers are not like picking up balls....u can pick the same no. second time....
[45]
odd no. 23
even no. 23
first one to be even 23/46
then second one to be odd 23/46(not 23/45...because the sample space does not change)
so probab 1/2*1/2=1/4
again
first one to be odd 23/46
then second one to be even 23/46...
so final ans 1/2
ee e
oo e
eo o
oe o
There are 23 even & 23 odd nos. out of the 46
Fav. case = 1 odd & 1 even = 23C1 * 23C1 = 23*23
Sample space = 46C2 = 23*45
P = (23*23)/(23*45) = 23/45