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




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Re: prob. example
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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
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14 Jan 2008, 05:15
3/8Marco 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^(nm) 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
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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
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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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Re: prob. example
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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



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Re: prob. example
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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
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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
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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
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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 \(\frac{\frac{P^3_3}{2!}}{2^3}=\frac{3}{8}\)



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Marco and Maria toss a coin three times. Each time a head is
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22 Jan 2012, 04:41
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=11+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: Marco and Maria toss a coin three times. Each time a head is
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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
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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
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16 Apr 2017, 20:13
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=11+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: Marco and Maria toss a coin three times. Each time a head is
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16 Apr 2017, 22:49
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=11+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
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28 Jul 2017, 10:27
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 $1 is for HHT scenario > Marco's balance=11+1=1. This scenario can occur in 3!/2! ways (# of permutations of 3 letters HHT out of which 2 H's are identical). As the total # of outcomes = 2^3 then P(HHT)=favorable/total=3/8. Would appreciate help with this! Does number of ways matter here? We are just concerned with the probability of Marco being 1$ 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
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29 Dec 2017, 07:38
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
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29 Dec 2017, 07:44
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
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10 Apr 2020, 19:52
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
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