Help me Out

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.

6 Answers

71
Vivek @ Born this Way ·

We 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
Aditya Bhutra ·

@vivek - the question asks for the number of ways , not the probability :P

71
Vivek @ Born this Way ·

LOL!!

By the way, How was aieee If you gave it offline. ? Give me some idea!

262
Aditya Bhutra ·

i opted for online .

71
Vivek @ Born this Way ·

Nice. Me too..!

7
Sigma ·

i think A will win once he wins more then n games. so it should be

2nCn+1 + 2nCn+2.....2n C 2n.

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