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John throws a coin until a series of three consecutive heads [#permalink]
 20 Oct 2008, 05:17
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Question Stats:
49% (01:46) correct
50% (00:40) wrong based on 65 sessions
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}Source: GMAT Club Tests - hardest GMAT questions
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Re: GMATClub M15: Probability [#permalink]
20 Oct 2008, 07:00
is d ans b....
1/16 + 1/16
2/16
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Re: GMATClub M15: Probability [#permalink]
20 Oct 2008, 07:26
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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]
20 Oct 2008, 10:51
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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nice explanation, got tricked by this Q. 
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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. 
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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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I said E.........DUMB MISTAKE!! I rushed it and forgot about the "consecutive" part.
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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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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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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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Yes B, got this one correct.
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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]
11 Dec 2012, 08:57
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, 23:41, edited 1 time in total.
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Re: John throws a coin until a series of three consecutive heads [#permalink]
11 Dec 2012, 20:58
very awesome question....got into the trap n selected 'D'...careful reading is so necessary.... Posted from my mobile device 
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Re: John throws a coin until a series of three consecutive heads
[#permalink]
11 Dec 2012, 20:58
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close
