Marco and Maria toss a coin three times. Each time a head is : GMAT Problem Solving (PS)
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# Marco and Maria toss a coin three times. Each time a head is

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

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14 Jan 2008, 04:57
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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. CEO Joined: 17 Nov 2007 Posts: 3589 Concentration: Entrepreneurship, Other Schools: Chicago (Booth) - Class of 2011 GMAT 1: 750 Q50 V40 Followers: 545 Kudos [?]: 3554 [1] , given: 360 Re: prob. example [#permalink] ### Show Tags 14 Jan 2008, 05:15 1 This post received KUDOS Expert's post 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
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14 Jan 2008, 06:50
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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. 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 CEO Joined: 29 Mar 2007 Posts: 2583 Followers: 19 Kudos [?]: 420 [1] , given: 0 Re: prob. example [#permalink] ### Show Tags 15 Jan 2008, 10:01 1 This post received KUDOS 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
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27 Jan 2008, 18:39
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
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25 Aug 2008, 08:05
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: 185 Followers: 2 Kudos [?]: 143 [0], given: 3 Re: prob. example [#permalink] ### Show Tags 27 Sep 2009, 05:50 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: 362 Followers: 11 Kudos [?]: 375 [0], given: 47 Re: prob. example [#permalink] ### Show Tags 14 Feb 2010, 14:18 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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14 Feb 2010, 23:16
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
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Re: Marco and Maria toss a coin three times. Each time a head is [#permalink]

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22 Jan 2012, 04:21
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 Joined: 02 Sep 2009 Posts: 36540 Followers: 7072 Kudos [?]: 93038 [1] , given: 10541 Re: Marco and Maria toss a coin three times. Each time a head is [#permalink] ### Show Tags 22 Jan 2012, 04:41 1 This post received KUDOS Expert's post 1 This post was BOOKMARKED 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.
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Re: Marco and Maria toss a coin three times. Each time a head is [#permalink]

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08 Dec 2013, 09:37
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Re: Marco and Maria toss a coin three times. Each time a head is [#permalink]

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12 Sep 2014, 20:27
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
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Marco and Maria toss a coin three times. Each time a head is [#permalink]

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15 Jun 2015, 21:46
probability of wining=1/2
probability of loosing=1/2

So,
1/2*1/2*1/2*3c2=3/8
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Re: Marco and Maria toss a coin three times. Each time a head is [#permalink]

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15 Jun 2015, 21:51
bunnel i have one confusion its 3P2 or 3c2
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Re: Marco and Maria toss a coin three times. Each time a head is [#permalink]

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29 Jun 2016, 20:27
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Re: Marco and Maria toss a coin three times. Each time a head is   [#permalink] 29 Jun 2016, 20:27
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