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VP  Joined: 22 Nov 2007
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Marco and Maria toss a coin three times. Each time a head is  [#permalink]

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Marco and Maria toss a coin three times. Each time a head is obtained, Marco gives Maria $1. Each time a tail is obtained, Maria gives Marco$1. What is the probability that after the 3 tosses, Marco has $1 less than he did before the 3 tosses? it is an example with coins, without possible answer choices. please provide explanations. Most Helpful Community Reply Director  Joined: 03 Sep 2006 Posts: 744 Re: prob. example [#permalink] Show Tags 3 3 marcodonzelli wrote: Marco and Maria toss a coin three times. Each time a head is obtained, Marco gives Maria$1. Each time a tail is obtained, Maria gives Marco $1. What is the probability that after the 3 tosses, Marco has$1 less than he did before the 3 tosses?

it is an example with coins, without possible answer choices. please provide explanations.

Marco can have 1 dollar less than he had before the 3 tosses only in one case, i.e., when there is "H" twice and "T" once.

THH = (1/2)^3
HTH = (1/2)^3
HHT = (1/2)^3

THH + HTH + HHT = 3/8
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CEO  B
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3/8

Marco has $1 less than he did before the 3 tosses if we have 2 heads and 1 tail: we use formula: p=nCm*p^m*q^(n-m) where p - probability of a head q - probability of a tail m - the number of heads n - the total number of tosses. p=3C2*(1/2)^2*(1/2)=3/8 _________________ HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | Limited GMAT/GRE Math tutoring in Chicago SVP  Joined: 29 Mar 2007 Posts: 2328 Re: prob. example [#permalink] Show Tags 1 marcodonzelli wrote: Marco and Maria toss a coin three times. Each time a head is obtained, Marco gives Maria$1. Each time a tail is obtained, Maria gives Marco $1. What is the probability that after the 3 tosses, Marco has$1 less than he did before the 3 tosses?

it is an example with coins, without possible answer choices. please provide explanations.

We need a WLL scencario. it will be 1/2*1/2*1/2 --> 1/8

Now WLL --> 3!/2!1! --> 3 - the number of possible arrangements of WLL. So 3*1/8 -> 3/8
VP  Joined: 28 Dec 2005
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for him to end up with a buck less means that the outcome was two heads and one tail. And this can happen in any order.

So, using binomial theorem, we get: (3C2)*(1/2)^2*(1/2) = 3*1/8 = 3/8
SVP  Joined: 07 Nov 2007
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marcodonzelli wrote:
Marco and Maria toss a coin three times. Each time a head is obtained, Marco gives Maria $1. Each time a tail is obtained, Maria gives Marco$1. What is the probability that after the 3 tosses, Marco has $1 less than he did before the 3 tosses? it is an example with coins, without possible answer choices. please provide explanations. It should be two tails and 1 head. 3 WAYS possible = HTT+TTH+THT PROBABILITY = 3/ 2^3=3/8 _________________ Your attitude determines your altitude Smiling wins more friends than frowning Manager  Joined: 27 Oct 2008 Posts: 165 Re: prob. example [#permalink] Show Tags Marco and Maria toss a coin three times. Each time a head is obtained, Marco gives Maria$1. Each time a tail is obtained, Maria gives Marco $1. What is the probability that after the 3 tosses, Marco has$1 less than he did before the 3 tosses?

Soln: For Marco to lose one dollar, the possible events are (HHT,HTH,THH) = 3 ways
Total number of possibilites is = 2 * 2 * 2 = 8 ways

Thus probability that Marco loses 1$is = 3/8 Senior Manager  Joined: 22 Dec 2009 Posts: 291 Re: prob. example [#permalink] Show Tags marcodonzelli wrote: Marco and Maria toss a coin three times. Each time a head is obtained, Marco gives Maria$1. Each time a tail is obtained, Maria gives Marco $1. What is the probability that after the 3 tosses, Marco has$1 less than he did before the 3 tosses?

it is an example with coins, without possible answer choices. please provide explanations.

Can only happen.. if we have two Heads and one Tails....

HHT = (1/2)^3 x 3!/2! (Arrange HTT) = 3/8
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Manager  Joined: 01 Feb 2010
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marcodonzelli wrote:
Marco and Maria toss a coin three times. Each time a head is obtained, Marco gives Maria $1. Each time a tail is obtained, Maria gives Marco$1. What is the probability that after the 3 tosses, Marco has $1 less than he did before the 3 tosses? it is an example with coins, without possible answer choices. please provide explanations. For Marco to have 1$ less than he did before the 3 tosses so 2 heads and one tail have to come.
p(event) = p(h)*p(h)*p(t) + p(h)*p(t)*p(h) + p(t)*p(h)*p(h) = (1/2)^3 + (1/2)^3 + (1/2)^3 = 3/8
Intern  Joined: 06 Jul 2011
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Re: Marco and Maria toss a coin three times. Each time a head is  [#permalink]

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A combinatoric approach:

There are $$2^3$$ total permutations of possible results.

The ones where he ends up with -1$obviously consist of 2x head, 1x tails. We can calculate the number of ways this can happen (head, head, tails or head, tail, head etc.) with the formula $$\frac{P^3_3}{2!} = \frac{3!}{2!} = 3$$ ($$P^3_3$$ is the number of permutation of 3 different items. However, since 2 items are the same (head & head), we need to divide by the factorial of the number of equal items, so $$2!$$) This gives us a probabilitiy of $$\frac{\frac{P^3_3}{2!}}{2^3}=\frac{3}{8}$$ Math Expert V Joined: 02 Sep 2009 Posts: 55718 Marco and Maria toss a coin three times. Each time a head is [#permalink] Show Tags 1 2 marcodonzelli wrote: Marco and Maria toss a coin three times. Each time a head is obtained, Marco gives Maria$1. Each time a tail is obtained, Maria gives Marco $1. What is the probability that after the 3 tosses, Marco has$1 less than he did before the 3 tosses?

it is an example with coins, without possible answer choices. please provide explanations.

No need to complicate. The only way Marco to loose $1 is for HHT scenario --> Marco's balance=-1-1+1=-1. This scenario can occur in 3!/2! ways (# of permutations of 3 letters HHT out of which 2 H's are identical). As the total # of outcomes = 2^3 then P(HHT)=favorable/total=3/8. _________________ Manager  Joined: 28 Aug 2013 Posts: 75 Location: India Concentration: Operations, Marketing Schools: Insead '14, ISB '15 GMAT Date: 08-28-2014 GPA: 3.86 WE: Supply Chain Management (Manufacturing) Re: Marco and Maria toss a coin three times. Each time a head is [#permalink] Show Tags Its 3/8 Probability of individual event : 1/2 No. of individual events : 3 therefore 1/2 X 1/2 X 1/2 = 1/8 No. of ways in which we can contain favorable result : 3 they are HTT, THT, TTH thus 3X1/8 = 3/8 _________________ G-prep1 540 --> Kaplan 580-->Veritas 640-->MGMAT 590 -->MGMAT 2 640 --> MGMAT 3 640 ---> MGMAT 4 650 -->MGMAT 5 680 -- >GMAT prep 1 570 Give your best shot...rest leave upto Mahadev, he is the extractor of all negativity in the world !! Intern  Joined: 14 Apr 2015 Posts: 5 Concentration: Human Resources, Technology GPA: 3.5 WE: Information Technology (Computer Software) Marco and Maria toss a coin three times. Each time a head is [#permalink] Show Tags probability of wining=1/2 probability of loosing=1/2 So, 1/2*1/2*1/2*3c2=3/8 _________________ Regards, YS (I can,I WIL!!!) Intern  S Joined: 26 Feb 2017 Posts: 29 Location: Brazil GMAT 1: 610 Q45 V28 GPA: 3.11 Re: Marco and Maria toss a coin three times. Each time a head is [#permalink] Show Tags 1 Bunuel wrote: marcodonzelli wrote: Marco and Maria toss a coin three times. Each time a head is obtained, Marco gives Maria$1. Each time a tail is obtained, Maria gives Marco $1. What is the probability that after the 3 tosses, Marco has$1 less than he did before the 3 tosses?

it is an example with coins, without possible answer choices. please provide explanations.

No need to complicate. The only way Marco to loose $1 is for HHT scenario --> Marco's balance=-1-1+1=-1. This scenario can occur in 3!/2! ways (# of permutations of 3 letters HHT out of which 2 H's are identical). As the total # of outcomes = 2^8 then P(HHT)=favorable/total=3/8. Bunuel Should not be Total # of outcomes = 2^3 instead of 2^8 ?? Tks Math Expert V Joined: 02 Sep 2009 Posts: 55718 Re: Marco and Maria toss a coin three times. Each time a head is [#permalink] Show Tags vitorpteixeira wrote: Bunuel wrote: marcodonzelli wrote: Marco and Maria toss a coin three times. Each time a head is obtained, Marco gives Maria$1. Each time a tail is obtained, Maria gives Marco $1. What is the probability that after the 3 tosses, Marco has$1 less than he did before the 3 tosses?

it is an example with coins, without possible answer choices. please provide explanations.

No need to complicate. The only way Marco to loose $1 is for HHT scenario --> Marco's balance=-1-1+1=-1. This scenario can occur in 3!/2! ways (# of permutations of 3 letters HHT out of which 2 H's are identical). As the total # of outcomes = 2^8 then P(HHT)=favorable/total=3/8. Bunuel Should not be Total # of outcomes = 2^3 instead of 2^8 ?? Tks Sure. It's 2^3 = 8 NOT 2^8 = 8. Edited. Thank you. _________________ Current Student B Joined: 30 Jan 2017 Posts: 6 Location: India Concentration: General Management, Marketing GMAT 1: 650 Q47 V35 GPA: 4 WE: Account Management (Advertising and PR) Re: Marco and Maria toss a coin three times. Each time a head is [#permalink] Show Tags Bunuel wrote: marcodonzelli wrote: Marco and Maria toss a coin three times. Each time a head is obtained, Marco gives Maria$1. Each time a tail is obtained, Maria gives Marco $1. What is the probability that after the 3 tosses, Marco has$1 less than he did before the 3 tosses?

it is an example with coins, without possible answer choices. please provide explanations.

No need to complicate. The only way Marco to loose $1 is for HHT scenario --> Marco's balance=-1-1+1=-1. This scenario can occur in 3!/2! ways (# of permutations of 3 letters HHT out of which 2 H's are identical). As the total # of outcomes = 2^3 then P(HHT)=favorable/total=3/8. Would appreciate help with this! Does number of ways matter here? We are just concerned with the probability of Marco being -1$ and that's 1/8 right?
I'm not clear about why we need to consider the other variations of HHT when we are dealing with a money situation.

Thanks!
Intern  B
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Marco and Maria toss a coin three times. Each time a head is  [#permalink]

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Hi,

What is the approximate difficulty of this question on the 800 scale?

Thank you
CEO  V
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Re: Marco and Maria toss a coin three times. Each time a head is  [#permalink]

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Top Contributor
krikre wrote:
Hi,

What is the approximate difficulty of this question on the 800 scale?

Thank you

I'd say around 600.

Cheers,
Brent
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Re: Marco and Maria toss a coin three times. Each time a head is  [#permalink]

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