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Marco and Maria toss a coin three times. Each time a head is [#permalink]

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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.

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.

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.

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

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

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

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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|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

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

Re: Marco and Maria toss a coin three times. Each time a head is [#permalink]

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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 \(\frac{\frac{P^3_3}{2!}}{2^3}=\frac{3}{8}\)

Re: Marco and Maria toss a coin three times. Each time a head is [#permalink]

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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 = 2^8 then P(HHT)=favorable/total=3/8. _________________

Re: Marco and Maria toss a coin three times. Each time a head is [#permalink]

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08 Dec 2013, 10:37

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Re: Marco and Maria toss a coin three times. Each time a head is [#permalink]

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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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