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M15-31

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M15-31 [#permalink]

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If John throws a coin until a series of three consecutive heads or three consecutive tails appears, what is the probability that the game will end on the fourth throw?

A. \(\frac{1}{16}\)
B. \(\frac{2}{16}\)
C. \(\frac{3}{16}\)
D. \(\frac{4}{16}\)
E. \(\frac{6}{16}\)
[Reveal] Spoiler: OA

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New post 16 Sep 2014, 00:57
Official Solution:

If John throws a coin until a series of three consecutive heads or three consecutive tails appears, what is the probability that the game will end on the fourth throw?

A. \(\frac{1}{16}\)
B. \(\frac{2}{16}\)
C. \(\frac{3}{16}\)
D. \(\frac{4}{16}\)
E. \(\frac{6}{16}\)

Calculate the probability that either of the two sequences, \(THHH\) or \(HTTT\), will occur. The probability of \(THHH\) occurring = \((\frac{1}{2})^4 = \frac{1}{16}\). The probability of \(HTTT\) occurring = \((\frac{1}{2})^4 = \frac{1}{16}\). The probability of either \(THHH\) or \(HTTT\) occurring = \(\frac{1}{16} + \frac{1}{16} = \frac{2}{16}\).

Answer: B
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Collection of Questions:
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Re: M15-31 [#permalink]

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New post 24 Aug 2015, 04:09
Since there are two options, Beroulli's formula can also be applied:

\(\frac{2!}{1!1!}*\frac{1}{2}^{4-1}*\frac{1}{2}^{4-3}\)

\(2*\frac{1}{8}*\frac{1}{2}\)

\(\frac{2}{16}\)

B.

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Re: M15-31 [#permalink]

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New post 21 Oct 2015, 10:22
Bunuel wrote:
Official Solution:

If John throws a coin until a series of three consecutive heads or three consecutive tails appears, what is the probability that the game will end on the fourth throw?

A. \(\frac{1}{16}\)
B. \(\frac{2}{16}\)
C. \(\frac{3}{16}\)
D. \(\frac{4}{16}\)
E. \(\frac{6}{16}\)

Calculate the probability that either of the two sequences, \(THHH\) or \(HTTT\), will occur. The probability of \(THHH\) occurring = \((\frac{1}{2})^4 = \frac{1}{16}\). The probability of \(HTTT\) occurring = \((\frac{1}{2})^4 = \frac{1}{16}\). The probability of either \(THHH\) or \(HTTT\) occurring = \(\frac{1}{16} + \frac{1}{16} = \frac{2}{16}\).

Answer: B


Hi Bunuel,

Here's how i am approaching this question.

The question is asking - out of 4 throws, what's the probability of getting 3 consecutive heads or 3 consecutive tails.

Total # of arrangements = 2 * 2 * 2 * 2 = 16

Now, 3 consecutive heads can be like - THHH or HHHT
Similarly, 3 consecutive tails can be like - HTTT or TTTH.

So, total # of favourable outcomes = 2 + 2 = 4.

So, the probability = \(\frac{4}{16}\)

Can you please point to where i am wrong ?
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Re: M15-31 [#permalink]

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New post 21 Oct 2015, 11:27
vivekgautam1 wrote:
Bunuel wrote:
Official Solution:

If John throws a coin until a series of three consecutive heads or three consecutive tails appears, what is the probability that the game will end on the fourth throw?

A. \(\frac{1}{16}\)
B. \(\frac{2}{16}\)
C. \(\frac{3}{16}\)
D. \(\frac{4}{16}\)
E. \(\frac{6}{16}\)

Calculate the probability that either of the two sequences, \(THHH\) or \(HTTT\), will occur. The probability of \(THHH\) occurring = \((\frac{1}{2})^4 = \frac{1}{16}\). The probability of \(HTTT\) occurring = \((\frac{1}{2})^4 = \frac{1}{16}\). The probability of either \(THHH\) or \(HTTT\) occurring = \(\frac{1}{16} + \frac{1}{16} = \frac{2}{16}\).

Answer: B


Hi Bunuel,

Here's how i am approaching this question.

The question is asking - out of 4 throws, what's the probability of getting 3 consecutive heads or 3 consecutive tails.

Total # of arrangements = 2 * 2 * 2 * 2 = 16

Now, 3 consecutive heads can be like - THHH or HHHT
Similarly, 3 consecutive tails can be like - HTTT or TTTH.

So, total # of favourable outcomes = 2 + 2 = 4.

So, the probability = \(\frac{4}{16}\)

Can you please point to where i am wrong ?


HHHT and TTTH does not work because in these cases the game ends on third throw not on the fourth.

Hope it helps.
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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: M15-31 [#permalink]

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New post 21 Oct 2015, 12:20
Bunuel wrote:
vivekgautam1 wrote:
Bunuel wrote:
Official Solution:

If John throws a coin until a series of three consecutive heads or three consecutive tails appears, what is the probability that the game will end on the fourth throw?

A. \(\frac{1}{16}\)
B. \(\frac{2}{16}\)
C. \(\frac{3}{16}\)
D. \(\frac{4}{16}\)
E. \(\frac{6}{16}\)

Calculate the probability that either of the two sequences, \(THHH\) or \(HTTT\), will occur. The probability of \(THHH\) occurring = \((\frac{1}{2})^4 = \frac{1}{16}\). The probability of \(HTTT\) occurring = \((\frac{1}{2})^4 = \frac{1}{16}\). The probability of either \(THHH\) or \(HTTT\) occurring = \(\frac{1}{16} + \frac{1}{16} = \frac{2}{16}\).

Answer: B


Hi Bunuel,

Here's how i am approaching this question.

The question is asking - out of 4 throws, what's the probability of getting 3 consecutive heads or 3 consecutive tails.

Total # of arrangements = 2 * 2 * 2 * 2 = 16

Now, 3 consecutive heads can be like - THHH or HHHT
Similarly, 3 consecutive tails can be like - HTTT or TTTH.

So, total # of favourable outcomes = 2 + 2 = 4.

So, the probability = \(\frac{4}{16}\)

Can you please point to where i am wrong ?


HHHT and TTTH does not work because in these cases the game ends on third throw not on the fourth.

Hope it helps.


awesome !
i missed that.

Thanks Bunuel.
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M15-31 [#permalink]

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New post 20 May 2016, 16:25
Bunuel wrote:
If John throws a coin until a series of three consecutive heads or three consecutive tails appears, what is the probability that the game will end on the fourth throw?

A. \(\frac{1}{16}\)
B. \(\frac{2}{16}\)
C. \(\frac{3}{16}\)
D. \(\frac{4}{16}\)
E. \(\frac{6}{16}\)


Is this a valid approach or did I just happen to get lucky:

Chance that it's either heads or tails on the 1st throw is 1
Chance for the next 3 throws to be the opposite of the 1st (heads first then 3 tails or vice versa) is \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\)

Multiply both and it's \(\frac{1}{8}\) which is reduced form of \(\frac{2}{16}\)
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Re: M15-31 [#permalink]

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New post 20 May 2016, 20:07
redfield wrote:
Bunuel wrote:
If John throws a coin until a series of three consecutive heads or three consecutive tails appears, what is the probability that the game will end on the fourth throw?

A. \(\frac{1}{16}\)
B. \(\frac{2}{16}\)
C. \(\frac{3}{16}\)
D. \(\frac{4}{16}\)
E. \(\frac{6}{16}\)


Is this a valid approach or did I just happen to get lucky:

Chance that it's either heads or tails on the 1st throw is 1
Chance for the next 3 throws to be the opposite of the 1st (heads first then 3 tails or vice versa) is \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\)

Multiply both and it's \(\frac{1}{8}\) which is reduced form of \(\frac{2}{16}\)


Hi,
you are absolutely correct in the concept you have used...
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Re: M15-31 [#permalink]

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New post 20 May 2016, 20:33
chetan2u wrote:

Hi,
you are absolutely correct in the concept you have used...


Chetan2u I think we could fill a book of me asking silly math questions and you coming to my rescue, thank you yet again!
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Re: M15-31 [#permalink]

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New post 16 Sep 2017, 10:38
I took a slightly different approach.

1. What are the odds that the first two throws result in different outcomes (which is necessary): 1/2.

2. Then, what are the odds of each of the following two throws matching the second throw? 1/2 * 1/2.

Answer = 1/2 * 1/2* 1/2 = 1/8.

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Re: M15-31   [#permalink] 16 Sep 2017, 10:38
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