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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]  14 Jan 2008, 05:57
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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: 3591 Concentration: Entrepreneurship, Other Schools: Chicago (Booth) - Class of 2011 GMAT 1: 750 Q50 V40 Followers: 230 Kudos [?]: 1298 [1] , given: 346 Re: prob. example [#permalink] 14 Jan 2008, 06:15 1 This post received KUDOS 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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Re: prob. example [#permalink]  14 Jan 2008, 07:50
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: 2618 Followers: 13 Kudos [?]: 142 [0], given: 0 Re: prob. example [#permalink] 15 Jan 2008, 11:01 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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Re: prob. example [#permalink]  27 Jan 2008, 19:39
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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Re: prob. example [#permalink]  25 Aug 2008, 09:05
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: 188 Followers: 1 Kudos [?]: 42 [0], given: 3 Re: prob. example [#permalink] 27 Sep 2009, 06:50 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: 368 Followers: 9 Kudos [?]: 136 [0], given: 47 Re: prob. example [#permalink] 14 Feb 2010, 15:18 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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Re: prob. example [#permalink]  15 Feb 2010, 00:16
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]  22 Jan 2012, 05:21
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} GMAT Club team member Joined: 02 Sep 2009 Posts: 11506 Followers: 1791 Kudos [?]: 9517 [0], given: 826 Re: Marco and Maria toss a coin three times. Each time a head is [#permalink] 22 Jan 2012, 05:41 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] 22 Jan 2012, 05:41
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