Very Easyyy

rkrish...now plz ans this one....more simplified...

Out of 2 consecutive integers two are chosen...find the probability that their sum is odd

rkrissh ...ur "drona of mathematics" yaar.......

ya dats rite.

actually we can not pick 1 twice out of 1,2

thatz the mistake i did

thanx rkrish...

Ans : 1 [9][9][9]

wats da ans..........

it cant 1/2 else u wudnt ask

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

O #$%&!!!

yeah crrct rk

now is it ok.

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