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M21-11

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M21-11  [#permalink]

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New post 16 Sep 2014, 01:10
2
10
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

58% (01:13) correct 42% (01:24) wrong based on 126 sessions

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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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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.
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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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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

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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?
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Re: M21-11  [#permalink]

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New post 08 Aug 2019, 02:36
Probability of HTH in 4 chances is 1/2 * 1/2 * 1/2 * 1/2
The same can happen in 4C3 ways. = 4 * 1/16 = 1/4

Please advise where am i going wrong?
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Re: M21-11   [#permalink] 08 Aug 2019, 02:36
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