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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] New post 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.
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Re: prob. example [#permalink] New post 14 Jan 2008, 05:15
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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
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Re: prob. example [#permalink] New post 14 Jan 2008, 06:50
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
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Re: prob. example [#permalink] New post 15 Jan 2008, 10:01
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] New post 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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Re: prob. example [#permalink] New post 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
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Re: prob. example [#permalink] New post 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
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Re: prob. example [#permalink] New post 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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Re: prob. example [#permalink] New post 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] New post 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}
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Re: Marco and Maria toss a coin three times. Each time a head is [#permalink] New post 22 Jan 2012, 04:41
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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.


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