srikanth9502
In a certain game, what is the probability that Martha wins the first 5 rounds and loses the sixth?
Statement #1: The chance that Martha wins the first 4 rounds and loses the fifth is \(\tfrac{1}{32}\).
Statement #2: The chance that Martha wins the first 6 rounds and loses the seventh is \(\tfrac{1}{128}\).
Dear
srikanth9502,
OK, I found the correct fractions on another website. My friend, please be careful in posting anything here: please double-check that all parts of a problem have been correctly copied. In preparing for the GMAT, every detail matters. My friend, how you do anything is how you do everything.
What this question is trying to do is lull us into a pattern, thinking that it should be reciprocals of the powers of 2. If the game is something that gets twice as hard to win each time, then of course, the missing probability, the probability of winning the first 5 and losing the 6th, would be 1/64. That is the tempting mistake to make.
In fact, we know nothing about the nature of the game and how its rounds work. Perhaps the rule is that winning the 5th round automatically means that one wins the 6th round, so this probability would be zero. Perhaps, there's some other rule, and the probability is 10% or 85% or something else. We really have no idea at all.
OA = (E)
This question pulls a cheap trick and tries to fool students. This is not really a well thought out question. Here's a much better DS question:
Value of Square of BinomialDoes all this make sense?
Mike