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M15-31

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Math Expert
Joined: 02 Sep 2009
Posts: 43361

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15 Sep 2014, 23:56
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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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Math Expert
Joined: 02 Sep 2009
Posts: 43361

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15 Sep 2014, 23:57
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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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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.
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Joined: 14 Mar 2015
Posts: 35
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: 43361

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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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Posts: 35
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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: 294

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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: 5539

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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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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

BANGALORE/-

Senior Manager
Joined: 18 Aug 2014
Posts: 294

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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: 39

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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.
Intern
Joined: 26 Dec 2017
Posts: 10

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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.
Math Expert
Joined: 02 Sep 2009
Posts: 43361

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14 Jan 2018, 06:53
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Expert's post
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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15 Jan 2018, 20:44
Got it Thank you Bunuel for clarifying
Re: M15-31   [#permalink] 15 Jan 2018, 20:44
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