The question is not correct because there are contradictions between the statements 1 and 2 ; and statements 2 and 3, as labeled below:

Statement 1: Assume that player A needs while player B needs "more" dollars to win the game (both and are non-negative integers)

Statement 2: occurs if and only if there are at least heads in the fi…rst trials.

Statement 3: Then use the binomial distribution to write an expression for

In Statement 1, the word "more" should have been struck out.

P(A) in Statement 3 should have been

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Assuming A needs to win dollars, and B, dollars, A will occur, if and only if there are heads in the first trials.

Example: Say n=3 dollars, m=2 dollars. For A to win there are n+m-1 trials, which have n+m-1=4 possible sample spaces {HHH},{HHTH},{HTHH},{THHH}

Note: This is not a fair question because it does not tell us who will go first in the game. There is a big difference as to who will go first because the first to go will have a much better probability of winning. This you can test yourself by considering the expected value of each player. A flaw is found in this question in which it expected the solution to align with binomial distribution, for for which after a string of number of heads appeared, B would bother to complete the remaining number trials.