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M21-11

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M21-11  [#permalink]

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New post 16 Sep 2014, 01:10
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If a coin is tossed until the sequence \(HTH\) appears (\(H\) denotes the fall of heads, \(T\) denotes the fall of tails), what is the probability that the game will end with 4th throw?

A. \(\frac{1}{16}\)
B. \(\frac{3}{32}\)
C. \(\frac{1}{8}\)
D. \(\frac{3}{16}\)
E. \(\frac{1}{4}\)

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M21-11  [#permalink]

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New post 16 Sep 2014, 01:10
Official Solution:

If a coin is tossed until the sequence \(HTH\) appears (\(H\) denotes the fall of heads, \(T\) denotes the fall of tails), what is the probability that the game will end with 4th throw?

A. \(\frac{1}{16}\)
B. \(\frac{3}{32}\)
C. \(\frac{1}{8}\)
D. \(\frac{3}{16}\)
E. \(\frac{1}{4}\)

We look for either \(HHTH\) or \(THTH\) sequence. The probability of either is \((\frac{1}{2})^4 = \frac{1}{16}\). The combined probability is \(\frac{2}{16} = \frac{1}{8}\).

Answer: C
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Re: M21-11  [#permalink]

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New post 05 Oct 2014, 22:12
I think provided solution as per the language of this question is not right.

We are asked to find winning probability after the 4th throw. As per my understanding it means in sequence any throw after 4th attempt.

XXHTH
XXXXHTH
XXXXXXXHTH
XXXXXXXXXXXXHTH

I tried to find all winning cases within 4th throw.

1) HTH = P = 1/8
2) XHTH (X any turn) = P = 1/8

Probability of winning within 4 attempts = 2/8 = 1/4

Thus required probability of winning after the 4th throw = 1 - 1/4 = 3/4
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Re: M21-11  [#permalink]

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New post 06 Oct 2014, 00:10
PiyushK wrote:
I think provided solution as per the language of this question is not right.

We are asked to find winning probability after the 4th throw. As per my understanding it means in sequence any throw after 4th attempt.

XXHTH
XXXXHTH
XXXXXXXHTH
XXXXXXXXXXXXHTH

I tried to find all winning cases within 4th throw.

1) HTH = P = 1/8
2) XHTH (X any turn) = P = 1/8

Probability of winning within 4 attempts = 2/8 = 1/4

Thus required probability of winning after the 4th throw = 1 - 1/4 = 3/4


The game will end after the fourth throw means that the fourth throw should be the last in the game.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M21-11  [#permalink]

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New post 06 Oct 2014, 09:01
2
Bunuel wrote:
PiyushK wrote:
I think provided solution as per the language of this question is not right.

We are asked to find winning probability after the 4th throw. As per my understanding it means in sequence any throw after 4th attempt.

XXHTH
XXXXHTH
XXXXXXXHTH
XXXXXXXXXXXXHTH

I tried to find all winning cases within 4th throw.

1) HTH = P = 1/8
2) XHTH (X any turn) = P = 1/8

Probability of winning within 4 attempts = 2/8 = 1/4

Thus required probability of winning after the 4th throw = 1 - 1/4 = 3/4


The game will end after the fourth throw means that the fourth throw should be the last in the game.


Then sentence should be:
what is the probability that the game will end with 4th throw?
what is the probability that the game will end at fourth throw?

After 4th throw is ambiguous; I discussed this question with other people and they also inferred it same as I did.
_________________

Piyush K
-----------------------
Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison
Don't forget to press--> Kudos :)
My Articles: 1. WOULD: when to use? | 2. All GMATPrep RCs (New)
Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

Math Expert
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Re: M21-11  [#permalink]

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New post 06 Oct 2014, 10:01
1
PiyushK wrote:
Bunuel wrote:
PiyushK wrote:
I think provided solution as per the language of this question is not right.

We are asked to find winning probability after the 4th throw. As per my understanding it means in sequence any throw after 4th attempt.

XXHTH
XXXXHTH
XXXXXXXHTH
XXXXXXXXXXXXHTH

I tried to find all winning cases within 4th throw.

1) HTH = P = 1/8
2) XHTH (X any turn) = P = 1/8

Probability of winning within 4 attempts = 2/8 = 1/4

Thus required probability of winning after the 4th throw = 1 - 1/4 = 3/4


The game will end after the fourth throw means that the fourth throw should be the last in the game.


Then sentence should be:
what is the probability that the game will end with 4th throw?
what is the probability that the game will end at fourth throw?

After 4th throw is ambiguous; I discussed this question with other people and they also inferred it same as I did.


Than you, Edited as you've suggested:

If a coin is tossed until the sequence \(HTH\) appears (\(H\) denotes the fall of heads, \(T\) denotes the fall of tails), what is the probability that the game will end with 4th throw?
_________________

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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re M21-11  [#permalink]

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New post 14 Jul 2016, 02:08
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Though the answer is correct, I am wary of the explanation provided. In the solution, we are provided that there are 2 possible outcomes- THTH and HHTH out of the possible 16 outcomes. However, these 16 possible outcomes also include the cases HTHH and HTHT, which are not possible because the game would have ended with 3rd toss.

Shouldn't the solution be: [Probability that the game does not end with 3rd throw (7/8)] X [Probability that the game ends with 4th throw (2/14)]= 1/8
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Re: M21-11  [#permalink]

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New post 11 Nov 2016, 14:46
Basically it does not matter what we get in the first throw.
The last three throw should be HTH
There is only one way of getting this out of total 8 possibilities, so probability = 1/8
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Re: M21-11  [#permalink]

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New post 17 Sep 2018, 10:32
Bunuel wrote:
If a coin is tossed until the sequence \(HTH\) appears (\(H\) denotes the fall of heads, \(T\) denotes the fall of tails), what is the probability that the game will end with 4th throw?

A. \(\frac{1}{16}\)
B. \(\frac{3}{32}\)
C. \(\frac{1}{8}\)
D. \(\frac{3}{16}\)
E. \(\frac{1}{4}\)



Hi Bunuel or anyone else that could answer this:

given that the H in “HTH” could occur in two different ways, why isn’t the answer 1/4?

+1 for an explanation

Posted from my mobile device
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Re: M21-11  [#permalink]

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New post 17 Sep 2018, 20:23
1
TippingPoint93 wrote:
Bunuel wrote:
If a coin is tossed until the sequence \(HTH\) appears (\(H\) denotes the fall of heads, \(T\) denotes the fall of tails), what is the probability that the game will end with 4th throw?

A. \(\frac{1}{16}\)
B. \(\frac{3}{32}\)
C. \(\frac{1}{8}\)
D. \(\frac{3}{16}\)
E. \(\frac{1}{4}\)



Hi Bunuel or anyone else that could answer this:

given that the H in “HTH” could occur in two different ways, why isn’t the answer 1/4?

+1 for an explanation

Posted from my mobile device


What do you mean "H could occur in two different ways"? How are you getting 1/4?
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
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Re: M21-11  [#permalink]

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New post 17 Sep 2018, 20:54
Bunuel wrote:
TippingPoint93 wrote:
Bunuel wrote:
If a coin is tossed until the sequence \(HTH\) appears (\(H\) denotes the fall of heads, \(T\) denotes the fall of tails), what is the probability that the game will end with 4th throw?

A. \(\frac{1}{16}\)
B. \(\frac{3}{32}\)
C. \(\frac{1}{8}\)
D. \(\frac{3}{16}\)
E. \(\frac{1}{4}\)



Hi Bunuel or anyone else that could answer this:

given that the H in “HTH” could occur in two different ways, why isn’t the answer 1/4?

+1 for an explanation

Posted from my mobile device


What do you mean "H could occur in two different ways"? How are you getting 1/4?


Bunuel,

I was able to figure it out. I had made a silly mistake in my thought process. Thanks :)
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Re: M21-11 &nbs [#permalink] 17 Sep 2018, 20:54
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