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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.
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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^(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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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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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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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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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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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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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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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}
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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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