Home
Class 12
MATHS
2n boys are randomly divided into two su...

`2n` boys are randomly divided into two subgroups containint `n` boys each. The probability that eh two tallest boys are in different groups is `n//(2n-1)` b. `(n-1)//(2n-1)` c. `(n-1)//4n^2` d. none of these

A

`(n)/(2n-1)`

B

`(n-1)/(2n -1)`

C

`(2n -1)/(4n^(2))`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Similar Questions

Explore conceptually related problems

2n boys are randomly divided into two subgroups containing n boys each.The probability that the two tallest boys are in different groups is

If n toys are distributed among N boys randomly, then the probability that a particular boy gets r(lt n) toys is

A group of 2n boys and 2n girls is randomly divided into two equal groups.The probability that each group contains the same number of boys and girls is (A) 1/2 (B) 1/n (c) 1/2n (D) None of these

Suppose n(>=3) persons are sitting in a row. Two of them are selected at random.The probability that they are not together is (A) 1-(2)/(n) (B) (2)/(n-1) (C) 1-(1)/(n) (D) nonoe of these

A draws a card from a pack of n cards marked 1,2,...,n. The card is replaced in the pack and B draws a card.Then the probability that A draws a higher card than B is (n+1)2n b.1/2 c.(n-1)2n d.none of these

The total number of ways in which 2n persons can be divided into n couples is a.(2n!)/(n!n!) b.(2n!)/((2!)^(3)) C.(2n!)/(n!(2!)^(n)) d.none of these

n letters are written to n different persons and addresses on the n envelopes are also written. If the letters are placed in the envelopes at random,the probability that at least one letter is not placed in the right envelope,is (A) 1-(1)/(n) (B) 1-(1)/(2n) (C) 1-(1)/(n^(2))(D)1-(1)/(n!)

The number of terms in the expansion of (x+1/x+1)^n is (A) 2n (B) 2n+1 (C) 2n-1 (D) none of these

(lim_(n rarr oo)[(2n)/(2n^(2)-1)(cos(n+1))/(2n-1)-(n)/(1-2n)(n(-1)^(n))/(n^(2)+1)]is1(b)-1(c)0(d) none of these