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Rosalie tosses a coin a number of times and counts the tota

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Rosalie tosses a coin a number of times and counts the tota

by varun289 » Wed May 01, 2013 12:16 am
Rosalie tosses a coin a number of times and counts the total number of times the coin lands tail-side up. How many times does she toss the coin?

The chance that the coin lands tail-side up exactly once is between 0.1 and 0.2.
The chance that the coin lands head side up exactly once is the same as the chance that the coin lands tail-side up exactly once.
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by Matt@VeritasPrep » Wed May 01, 2013 10:10 pm
What a great question!

Assuming we have a fair coin (a safe assumption on the GMAT), statement (1) is SUFFICIENT.

First, let's find a formula for the odds of getting exactly one tails. Say we flip the coin twice. Our outcomes are HH, HT, TH, and TT, so there are exactly two ways of getting one tails out of four total results. Now say we flip it three times: our outcomes are HHH, HHT, HTH, THH, TTH, THT, HTT, and TTT, giving us three single-tails results out of a total of 8.

A pattern is starting to emerge: if we flip the coin n times, there are n ways of getting exactly one tails, and 2^n total results. There are n arrangements of exactly one tails because our string of results (represented as HHT, or whatever) has n places, exactly one of which is tails, so there are n spots in which to "place" the tails. There are 2^n total arrangements because we have two possible results on each flip.

With this in mind, statement (1) really tells us that 0.1 < n/(2^n) < 0.2. [spoiler]The only integer value of n for which that's true is n = 5, so we're good![/spoiler]

Statement (2) is INSUFFICIENT: this is actually true for ANY number of flips, since tails and heads are equally likely. This is helpful in a DS sort of way, however: [spoiler]this statement is actually so useless that the answer can only be A or E, so you have a 50/50 shot if you're making an inspired guess ... pretty fitting for a question that involves the flip of a coin![/spoiler]
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