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John throws a coin until a series of three consecutive heads [#permalink]
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?

Re: GMATClub M15: Probability [#permalink]
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]
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.

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.

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

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.

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

Re: John throws a coin until a series of three consecutive heads [#permalink]
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.