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Director
Joined: 14 Jan 2007
Posts: 704

A fair coin is tossed 5 times. What is the probability
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12 May 2007, 13:59
Question Stats:
100% (00:00) correct 0% (00:00) wrong based on 4 sessions
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A fair coin is tossed 5 times. What is the probability getting at least 3 heads on consecutive tosses?
a. 3/16
b. 1/4
c. 7/24
d. 5/16
e. 15/32 == Message from the GMAT Club Team == THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION. This discussion does not meet community quality standards. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



Senior Manager
Joined: 03 May 2007
Posts: 253

It a hard problem, but I know how to solve it (I just learned binomial formula)
Ok the first step is to find combinations 5C3= 5!/(53)!*3!=5*4/2=10
Second is to find the probability of all wanted events, 1/2^3 (for all three it will be 1/8)
Third is to find the probability of all unwanted events, 1/2^2 (1/4)
now you have to multiply all these numbers
10/8*4
the answer is 10/32 or 5/16
P.S. sorry for a lame explanation, but it's very hard to explain (check the attached document)



Senior Manager
Joined: 19 Sep 2004
Posts: 304

Hi you answer would have been correct if the problem would have been "Three Heads" but I think you overlooked Consecutive heads...so the asnwer is 3/16.
Saurabh Malpani
Sergey_is_cool wrote: It a hard problem, but I know how to solve it (I just learned binomial formula)
Ok the first step is to find combinations 5C3= 5!/(53)!*3!=5*4/2=10 Second is to find the probability of all wanted events, 1/2^3 (for all three it will be 1/8) Third is to find the probability of all unwanted events, 1/2^2 (1/4)
now you have to multiply all these numbers
10/8*4
the answer is 10/32 or 5/16
P.S. sorry for a lame explanation, but it's very hard to explain (check the attached document)



Director
Joined: 26 Feb 2006
Posts: 812

saurabhmalpani wrote: Hi you answer would have been correct if the problem would have been "Three Heads" but I think you overlooked Consecutive heads...so the asnwer is 3/16. Saurabh Malpani Sergey_is_cool wrote: It a hard problem, but I know how to solve it (I just learned binomial formula)
Ok the first step is to find combinations 5C3= 5!/(53)!*3!=5*4/2=10 Second is to find the probability of all wanted events, 1/2^3 (for all three it will be 1/8) Third is to find the probability of all unwanted events, 1/2^2 (1/4)
now you have to multiply all these numbers
10/8*4
the answer is 10/32 or 5/16
P.S. sorry for a lame explanation, but it's very hard to explain (check the attached document)
i got 1/8.
HHHHH
THHHH
TTHHH
TTHHH
I am not sure how you got 6?



Manager
Joined: 13 Feb 2007
Posts: 61

Coin Toss
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12 May 2007, 23:07
The question stem asks not just about consecutive heads, but ''at least three'' consecutive heads: this means, of course, the sum of the probobility of 3 consecutive heads, the probability of 4 consecutive heads, and the probability of 5 consecutive heads. SoHimalayan you have these possibilies: HHHTT, THHHT, TTHHH, HHHHT, THHHH, and HHHHH. This ends up resulting in a probability of 6/32 = 3/16.



Director
Joined: 14 Jan 2007
Posts: 704

clue  count all cases where we can get the three or more consecutive heads.



Director
Joined: 13 Mar 2007
Posts: 522
Schools: MIT Sloan

Answer  A
Probability of atleast 3 = prob 3 + prob 4 + prob 5
Probability of 3 consecutive heads = 3 x (1/2)^5
(1st,2nd,3rd or 2nd,3rd,4th or 3rd,4th,5th)
Probability of 4 cons. heads = 2 x (1/2)^5
(1st,2nd,3rd,4th or 2nd,3rd,4th,5th)
Probability of 5 cons. heads = 1 x (1/2)^5
Total probability = 6 x (1/2)^5 = 3/16
Is there any direct formula to calculate the above ?
say probability of atleast 49 consecutive heads in 200 tosses .. ?



Director
Joined: 14 Jan 2007
Posts: 704

OA is 'B'.
Total ways = 2^5 =32
Ways of getting at least 3 consecutive heads are:8
HHHTT
HHHHT
HHHHH
TTHHH
THHHT
THHHH
HTHHH
HHHTH
hence probability = 8/32 =1/4



Senior Manager
Joined: 19 Sep 2004
Posts: 304

Oh **** missed two cases!!!
Thanks for pointing out the mistake.
Saurabh Malpani
vshaunak@gmail.com wrote: OA is 'B'.
Total ways = 2^5 =32 Ways of getting at least 3 consecutive heads are:8 HHHTT HHHHT HHHHH TTHHH THHHT THHHH HTHHH HHHTH
hence probability = 8/32 =1/4 == Message from the GMAT Club Team == THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION. This discussion does not meet community quality standards. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



NonHuman User
Joined: 09 Sep 2013
Posts: 11396

Re: A fair coin is tossed 5 times. What is the probability
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10 Apr 2018, 02:09
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Re: A fair coin is tossed 5 times. What is the probability
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10 Apr 2018, 02:09






