1) 2 players A and B play a series of 2n games. Each game can result in either a win or a loss for A. Find the total no. of ways in which A can win the series of these games. (All the games are to be played)
Ans: 22n-1 - 1/2. 2nCn.
71We have here 2n Bernoulli trials with probability of success = 1/2 and failure = 1/2
So P(A wins) = 2nCn+1 (1/2)2n + 2nCn+2 (1/2)2n + .... + 2nC2n (1/2)2n
Which comes out as P(A wins) = 1/2 - 2nCn22n+1 .
NOTE: There might be some calculation mistake. However the answer given by you is incorrect. See for n =1, it gives P = 1 which is sure event and not possible. The actual value is 1/4 which is given by the answer above.
262@vivek - the question asks for the number of ways , not the probability :P
71LOL!!
By the way, How was aieee If you gave it offline. ? Give me some idea!
7i think A will win once he wins more then n games. so it should be
2nCn+1 + 2nCn+2.....2n C 2n.