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John throws a coin until a series of three consecutive heads

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John throws a coin until a series of three consecutive heads [#permalink] New post 20 Oct 2008, 04:17
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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 after 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
B

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Re: GMATClub M15: Probability [#permalink] New post 20 Oct 2008, 06:00
is d ans b....

1/16 + 1/16

2/16
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Re: GMATClub M15: Probability [#permalink] New post 20 Oct 2008, 06:26
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scthakur wrote:
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 after the fourth throw?
a. 1/16
b. 2/16
c. 3/16
d. 4/16
e. 6/16

Correct answer is B.


Tricky question!! good one.

this trap for 4/16 (answer D).

TOTAL no. of ways. = 2^4=16

At the first look we may thought all below four combinations are correct.
HTTT
THHH
TTTH -- If this were the case game would have been over after 3 throws.
HHHT-- If this were the case game would have been over after 3 throws.


so only two possiblilites...
ans = 2/16
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Re: GMATClub M15: Probability [#permalink] New post 20 Oct 2008, 09:51
x2suresh wrote:
scthakur wrote:
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 after the fourth throw?
a. 1/16
b. 2/16
c. 3/16
d. 4/16
e. 6/16

Correct answer is B.


Tricky question!! good one.

this trap for 4/16 (answer D).

TOTAL no. of ways. = 2^4=16

At the first look we may thought all below four combinations are correct.
HTTT
THHH
[color=#BF0000]TTTH -- If this were the case game would have been over after 3 throws.
HHHT-- If this were the case game would have been over after 3 throws.[/color]

so only two possiblilites...
ans = 2/16



Nice explanation Suresh. I could not imagine the highlighted part and ended up selecting 4/16 as the answer.
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Re: M15 #31 [#permalink] New post 22 Sep 2009, 07:26
nice explanation, got tricked by this Q. :)
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Re: M15 #31 [#permalink] New post 30 Nov 2009, 13:33
ohh snapp this is a tricky one. I got the right answer in abt 15 sec but it took me over a minute to realize the trap of 4/16. i went over my answer abt 5 times thinking it was wrong. i really have to spend more time on looking at answer choices before i start cramping the maths. :x
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Re: M15 #31 [#permalink] New post 01 Dec 2009, 09:07
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Re: M15 #31 [#permalink] New post 05 Dec 2010, 12:16
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My solution:
1st throw - we don't care tail of head. Prob is 1
2nd -4th throws - we need 3 consequitive items, which are opposite to item in 1st throw. Therefore, cumulative probability is 1/2*1/2*1/2 = 1/8

It took me 20 sec to find the correct asnwer which is 2/16= 1/8
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Re: M15 #31 [#permalink] New post 07 Dec 2010, 08:15
I said E.........DUMB MISTAKE!! I rushed it and forgot about the "consecutive" part.
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Re: M15 #31 [#permalink] New post 07 Dec 2010, 14:34
Vorskl wrote:
My solution:
1st throw - we don't care tail of head. Prob is 1
2nd -4th throws - we need 3 consequitive items, which are opposite to item in 1st throw. Therefore, cumulative probability is 1/2*1/2*1/2 = 1/8

It took me 20 sec to find the correct asnwer which is 2/16= 1/8


Your reasoning is wrong since you do care what is the first throw - if it is H and then you get 2 more Hs the game ends. You got the correct answer by chance. Though, chance is also a factor in this test.

EDIT: I now see that you wrote: "which are opposite to item in 1st throw." so i guess it is correct.
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Re: M15 #31 [#permalink] New post 08 Dec 2010, 07:14
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B.

Two possibilities for this to occur:
THHH
HTTT

For both the prob of each event needs to be multiplied as they are dependent events:
Hence 1/2 * 1/2 * 1/2 * 1/2
=1/16
for both events multiply by 2 =>
2 * 1/16 = 2/16
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Re: M15 #31 [#permalink] New post 09 Dec 2011, 07:52
B.

all laid it out it would be 1*(1-1/2)*1/2*1/2 which equals 1/8 which equals 2/16. should take 40 seconds MAX once you understand the reasoning
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Re: M15 #31 [#permalink] New post 10 Dec 2011, 23:26
Yes B, got this one correct.
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Re: M15 #31 [#permalink] New post 11 Dec 2011, 15:48
Read the question wrong initially so I did by the 4th throw. :(
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Re: John throws a coin until a series of three consecutive heads [#permalink] New post 11 Dec 2012, 07:57
Got the Q and solution of this one.

Consecutive makes it a bit easy with only 2 desired cases.
Just wanted to understand the solution in case "consecutive" was not mentioned.

I think it will be an anagram in that case.

Favourable cases: ((For HHHT=: 4!/3!)+(For TTTH=:4!/3!))

Total cases= 4!

Hence Answer= (1/3)

Can somebody confirm or point out mistakes in the modified Q (in case consecutive is not mentioned)??

Thanks

Last edited by soumens on 11 Dec 2012, 22:41, edited 1 time in total.
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Re: John throws a coin until a series of three consecutive heads [#permalink] New post 11 Dec 2012, 19:58
very awesome question....got into the trap n selected 'D'...careful reading is so necessary....:(

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Re: John throws a coin until a series of three consecutive heads [#permalink] New post 05 Dec 2013, 11:35
My problem with this question is the way that it is worded. The phrase "What is the probability that the game ends after the fourth toss" could be interpreted as meaning what is the probabaility that the game does not end any time before the fifth throw, or in other words what is the probability that the game does not end on the fourth toss or before. This was my initial understanding, and although I eventually settled on what i assumed was the correct meaning (game ends right at fourth toss), I think this stem needs to be ammended to say "probability that the game ends immediately after the fourth throw" in order to be airtight.

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Re: John throws a coin until a series of three consecutive heads [#permalink] New post 05 Dec 2013, 11:43
Using simple counting:

Possible outcomes (Green--> correct outcome, red --> false outcome): 0 represents tail, 1 represents head

0000
0001

0010
0011
0100
0101
0110
0111
1000

1001
1010
1011
1100
1101
1110
1111


probability = (correct outcome) / possible outcome

P = 2/16
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Re: John throws a coin until a series of three consecutive heads [#permalink] New post 05 Dec 2013, 21:48
Excellent question...marked D on first attempt, but now understood why it is B.

1. THHH - 1/2 * 1/2 * 1/2 * 1/2 = 1/16
2. HTTT - 1/2 * 1/2 * 1/2 * 1/2 = 1/16

Adding both since it is either the first case or the second case.
So 1/16 + 1/16 = 2/16

So, B
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Re: John throws a coin until a series of three consecutive heads [#permalink] New post 07 Dec 2013, 09:48
ans is B.
since the coin is tossed 4 times there can be 16 total outcomes.
there are only 2 outcomes that satisfy the condition.
HTTT and THHH.
Re: John throws a coin until a series of three consecutive heads   [#permalink] 07 Dec 2013, 09:48
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