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Marco and Maria toss a coin three times. Each time a head is

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marcodonzelli
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Marco and Maria toss a coin three times. Each time a head is [#permalink] New post   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.
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Re: prob. example [#permalink] New post   14 Jan 2008, 07:50
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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
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Re: prob. example [#permalink] New post   14 Jan 2008, 06:15
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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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Marco and Maria toss a coin three times. Each time a head is [#permalink] New post   22 Jan 2012, 05:41
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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 = 2^3 then P(HHT)=favorable/total=3/8.
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Re: prob. example [#permalink] New post   15 Jan 2008, 11:01
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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.


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 Sep 2009, 06:50
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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?


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: Marco and Maria toss a coin three times. Each time a head is [#permalink] New post   22 Jan 2012, 05:21
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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
\(\frac{\frac{P^3_3}{2!}}{2^3}=\frac{3}{8}\)
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Re: Marco and Maria toss a coin three times. Each time a head is [#permalink] New post   16 Apr 2017, 21:13
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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
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Re: prob. example [#permalink] New post   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] New post   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
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Re: prob. example [#permalink] New post   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] New post   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] New post   12 Sep 2014, 21:27
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] New post   15 Jun 2015, 22:46
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] New post   16 Apr 2017, 23:49
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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.
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Re: Marco and Maria toss a coin three times. Each time a head is [#permalink] New post   28 Jul 2017, 11:27
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!
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Marco and Maria toss a coin three times. Each time a head is [#permalink] New post   29 Dec 2017, 08:38
Hi,

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

Thank you
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Re: Marco and Maria toss a coin three times. Each time a head is [#permalink] New post   29 Dec 2017, 08:44
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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] New post   10 Apr 2020, 20:52
Marco will be $1 less when she has lost 2 chances and won 1 chances.
Possible cases for Marco
WLL + LWL + LLW
Every time Probability is 1/2×1/2×1/2 =1/8
Therefore answer is 3/8

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Re: Marco and Maria toss a coin three times. Each time a head is [#permalink] New post   05 Nov 2022, 02:17
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Re: Marco and Maria toss a coin three times. Each time a head is [#permalink]
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