hi varsha,

we hav to arrange n persons along  d round table.

toatal ways of arranging them round d table = (n-1)!

now we wish 2 persons come together den we can do one thing dat we combine both of dem as a single person n after dat we hav (n-1) persons n

total ways of arranging dem = (n-1-1)! * 2! [ as dose 2 persons can be arranged demselves]

                                         = (n-2)! * 2

probability of this event = (n-2)! * 2 / (n-1)! = 2/ (n-1)

n odd against dis event = 1-2/(n-1) = (n-3)/(n-1).

hi varsha,

since out of 51 cards we checked 13 n all of dem r red, dat means out of d 38 cards, either of d two case may occur, 1.) one red card is missing or 2.) one black is missing

in 1st case, der wud be 26 blacks n 12 red

in 2nd case., der wud be 25 blacks n 13red

in total of 39 cards , der r 38 available n 1 is missing ,

required p = when black is missing/either red or black is missing

p (black is missing) = 26 c 25 * 13 c 13 /( 26 c 25 * 13c 13 + 26c16 * 13 c 12) = dis will give u

p = 2/3