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# M15-31

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Math Expert
Joined: 02 Sep 2009
Posts: 51261

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15 Sep 2014, 23:56
1
3
00:00

Difficulty:

45% (medium)

Question Stats:

59% (00:57) correct 41% (01:13) wrong based on 109 sessions

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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. $$\frac{6}{16}$$

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Math Expert
Joined: 02 Sep 2009
Posts: 51261

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15 Sep 2014, 23:57
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 = $$\frac{1}{16} + \frac{1}{16} = \frac{2}{16}$$.

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Joined: 15 Sep 2011
Posts: 326
Location: United States
WE: Corporate Finance (Manufacturing)

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24 Aug 2015, 03:09
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.
Intern
Joined: 14 Mar 2015
Posts: 33
Schools: ISB '18

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21 Oct 2015, 09:22
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 = $$\frac{1}{16} + \frac{1}{16} = \frac{2}{16}$$.

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 ?
_________________

It ain’t about how hard you hit. It’s about how hard you can get hit and keep moving forward;
how much you can take and keep moving forward.

That’s how winning is done!

Math Expert
Joined: 02 Sep 2009
Posts: 51261

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21 Oct 2015, 10:27
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 = $$\frac{1}{16} + \frac{1}{16} = \frac{2}{16}$$.

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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Intern
Joined: 14 Mar 2015
Posts: 33
Schools: ISB '18

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21 Oct 2015, 11:20
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 = $$\frac{1}{16} + \frac{1}{16} = \frac{2}{16}$$.

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.
_________________

It ain’t about how hard you hit. It’s about how hard you can get hit and keep moving forward;
how much you can take and keep moving forward.

That’s how winning is done!

Senior Manager
Joined: 18 Aug 2014
Posts: 324

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20 May 2016, 15:25
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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Math Expert
Joined: 02 Aug 2009
Posts: 7107

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20 May 2016, 19:07
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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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Senior Manager
Joined: 18 Aug 2014
Posts: 324

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20 May 2016, 19:33
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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Intern
Joined: 20 Aug 2017
Posts: 38
Location: United States (FL)
Schools: Stanford '20 (D)
GMAT 1: 750 Q49 V42
GPA: 3.4

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16 Sep 2017, 09:38
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.
Manager
Joined: 26 Dec 2017
Posts: 157

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14 Jan 2018, 06:07
Hi Experts,

I agree that the favorable outcomes are htttt or thhhh and total is 16.
I have done favorable calculation as (4!/3!)*2 and d/r is 16 which comes as 8/16.
I was wondering where I went wrong.
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Math Expert
Joined: 02 Sep 2009
Posts: 51261

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14 Jan 2018, 06:53
1
tejyr wrote:
Hi Experts,

I agree that the favorable outcomes are htttt or thhhh and total is 16.
I have done favorable calculation as (4!/3!)*2 and d/r is 16 which comes as 8/16.
I was wondering where I went wrong.

The game ends when three consecutive heads or three consecutive tails appears. So, the game to end on the fourth throw, we should have first T and then three H's or first H and then three T's. Thus, there are only two favorable cases THHH and HTTT (no need for factorial correction which accounts for different arrangements).
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Joined: 26 Dec 2017
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15 Jan 2018, 20:44
Got it Thank you Bunuel for clarifying
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27 Nov 2018, 06:47
Re: M15-31 &nbs [#permalink] 27 Nov 2018, 06:47
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# M15-31

Moderators: chetan2u, Bunuel

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