In a game a coin is tossed `2n+m` times and a player wins if he does not get any two consecutive outcomes same for at least `2n` times in a row. The probability that player wins the game is a. `(m+2)/(2^(2n)+1)` b. `(2n+2)/(2^(2n))` c. `(2n+2)/(2^(2n+1))` d. `(m+2)/(2^(2n))`

A coin is tossed (m+n) times with m>n. Show that the probability of getting m consecutive heads is (n+2)/2^(m+1)

Evaluate lim_(ntooo) (1)/(1+n^(2))+(2)/(2+n^(2))+...+(n)/(n+n^(2)).

Two players P_1a n dP_2 play a series of 2n games. Each game can result in either a win or a loss for P_1dot the total number of ways in which P_1 can win the series of these games is equal to a. 1/2(2^(2n)-^ "^(2n)C_n) b. 1/2(2^(2n)-2xx^"^(2n)C_n) c. 1/2(2^n-^"^(2n)C_n) d. 1/2(2^n-2xx^"^(2n)C_n)

A coin is tossed 2n times. The chance that the number of times one gets head is not equal to the number of times one gets tails is ((2n !))/((n !)^2)(1/2)^(2n) b. 1-((2n !))/((n !)^2) c. 1-((2n !))/((n !)^2)1/(4^n)^ d. none of these

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

(n+2)nC_0(2^(n+1))-(n+1)nC_1(2^(n))+(n)nC_2(2^(n-1))-.... is equal to

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

Find the sum of series upto n terms ((2n+1)/(2n-1))+3((2n+1)/(2n-1))^2+5((2n+1)/(2n-1))^3+...