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M21-11

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Math Expert
Joined: 02 Sep 2009
Posts: 52344

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16 Sep 2014, 00:10
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Difficulty:

45% (medium)

Question Stats:

54% (00:48) correct 46% (00:52) wrong based on 166 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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Posts: 52344

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16 Sep 2014, 00:10
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 $$\frac{2}{16} = \frac{1}{8}$$.

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05 Oct 2014, 21:12
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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05 Oct 2014, 23:10
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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06 Oct 2014, 08:01
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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Piyush K
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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
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Math Expert
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Posts: 52344

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06 Oct 2014, 09:01
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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14 Jul 2016, 01:08
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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Joined: 23 Apr 2016
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11 Nov 2016, 13:46
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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Joined: 06 Jul 2013
Posts: 54
Location: United States

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17 Sep 2018, 09:32
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
Math Expert
Joined: 02 Sep 2009
Posts: 52344

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17 Sep 2018, 19:23
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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17 Sep 2018, 19:54
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
Re: M21-11 &nbs [#permalink] 17 Sep 2018, 19:54
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