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A fair coin is tossed 6 times. What is the probability of getting no any two heads on consecutive tosses? a.21/64 b.42/64 c.19/64 d.19/42 e.31/64

Hi, there. I'm happy to help with this.

First of all, question is considerably harder and more pain-in-the-tush than what you will see on the GMAT. I don't know the source, but this seems to come from some over-achieving source that wants to give students questions much harder than the test.

Probability = (# of desired cases)/(total # of possible cases)

The denominator is very easy --- two possibilities for each toss, six tosses, so 2^6 = 64. That's the denominator.

For the numerator, we have to sort through cases:

Case One: six tails

For this case, obviously you can't have two heads in a row. There's only one way this can happen: TTTTTT ONE

Case Two: five tails, one head Again, it's impossible to have two heads in a row, because there's only one. There are six ways this could happen --- the H could occupy any of the six positions (HTTTTT, THTTTT, TTHTTT, TTTHTT, TTTTHT, and TTTTTH) SIX

Case Three: four tails, two heads This is the tricky case. There are 6C2 = 15 places that the two H's could land, but five of those (HHTTTT, THHTTT, TTHHTT, TTTHHT, and TTTTHH) involve the pair of H's together, which is forbidden. Excluding those five forbidden cases, we are left with 15 - 5 = 10 possibilities here. TEN

Case Four: three tails, three heads Now, things are starting to get crowded. We have to space our three H's out, with T's between them, so none of the H's touch. That leaves only two possibilities: HTHTHT and THTHTH. That's it: any other configuration would have two H's next to each other, which is forbidden. TWO

Re: A fair coin is tossed 6 times. What is the probability of getting no [#permalink]

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24 Sep 2012, 22:57

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A There are 64 possible outcomes. - If no head => 1 outcome - If we can only toss head once then there are 6 desired outcomes (eg: HTTTTT, THTTTT....) - If we can toss head twice then there are \(\frac{6!}{2!*4!}-5 = 10\) desired outcomes. - If we can toss head three times then there are 4 outcomes THTHTH, HTHTHT, HTTHTH, HTHTTH - 4, 5, 6 times -> no outcome => Probability = \(\frac{21}{64}\)

Last edited by monsama on 24 Sep 2012, 23:02, edited 3 times in total.

Re: A fair coin is tossed 6 times. What is the probability of getting no [#permalink]

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24 Sep 2012, 22:59

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mikemcgarry wrote:

saikarthikreddy wrote:

A fair coin is tossed 6 times. What is the probability of getting no any two heads on consecutive tosses? a.21/64 b.42/64 c.19/64 d.19/42 e.31/64

Case Four: three tails, three heads Now, things are starting to get crowded. We have to space our three H's out, with T's between them, so none of the H's touch. That leaves only two possibilities: HTHTHT and THTHTH. That's it: any other configuration would have two H's next to each other, which is forbidden. TWO

@mikemcgarry: in case 4, what about HTTHTH and HTHTTH => there are 4 possibilities :D

Re: A fair coin is tossed 6 times. What is the probability of getting no [#permalink]

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15 Aug 2015, 05:26

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Bunuel, is there any other way than counting method to solve this question?

Thanks.

Dear Swaroopdev I'm happy to respond.

My friend, a couple things to keep in mind. When Bunuel and the other experts show a solution to a problem, a relatively long and complicated solution, it's not as if we are making it complicated just for our own amusement. If there were a quick, easy, formulaic way to approach the problem, of course we would show that. In general, I would say that you can trust Bunuel and the other experts to show you the easiest, the most straightforward, and most efficient solution to any problem. Saying, "This solution looks hard. I don't like it. Can you show me an easier way?" is not the path that leads to excellence. The path that leads to excellence is all about challenging oneself to dive into what is most difficult and confusing. Assume that the path that Bunuel shows you is the optimal solution, and do your best to understand every last detail of it.

Also, there's something important thing to keep in mind about Counting and about Probability. Other branches of math, such as algebra, tend to be more formulaic and recipe based. If I give you a simple algebraic equation and ask you to solve for x, there's an easy recipe to follow. Counting and Probability are not primarily formula-based or recipe-based. Yes, there are a few formulas, but it is far from straightforward to know exactly when they can or can't be applied. Many Counting & Probability problems are about seeing the problem correct, interpreting the given information in a way that allows you to dissect the problem. Many students get into a "what should I do?" mode in problem solving, and with both Counting & Probability, it's important at the beginning of a problem to be not in the "what should I do?" mode, but in the "how do I look at this?" mode. When you are looking at the problem in the right way, what to do become obvious.

Bunuel, is there any other way than counting method to solve this question?

Thanks.

Dear Swaroopdev I'm happy to respond.

My friend, a couple things to keep in mind. When Bunuel and the other experts show a solution to a problem, a relatively long and complicated solution, it's not as if we are making it complicated just for our own amusement. If there were a quick, easy, formulaic way to approach the problem, of course we would show that. In general, I would say that you can trust Bunuel and the other experts to show you the easiest, the most straightforward, and most efficient solution to any problem. Saying, "This solution looks hard. I don't like it. Can you show me an easier way?" is not the path that leads to excellence. The path that leads to excellence is all about challenging oneself to dive into what is most difficult and confusing. Assume that the path that Bunuel shows you is the optimal solution, and do your best to understand every last detail of it.

Also, there's something important thing to keep in mind about Counting and about Probability. Other branches of math, such as algebra, tend to be more formulaic and recipe based. If I give you a simple algebraic equation and ask you to solve for x, there's an easy recipe to follow. Counting and Probability are not primarily formula-based or recipe-based. Yes, there are a few formulas, but it is far from straightforward to know exactly when they can or can't be applied. Many Counting & Probability problems are about seeing the problem correct, interpreting the given information in a way that allows you to dissect the problem. Many students get into a "what should I do?" mode in problem solving, and with both Counting & Probability, it's important at the beginning of a problem to be not in the "what should I do?" mode, but in the "how do I look at this?" mode. When you are looking at the problem in the right way, what to do become obvious.

Thank you for your response. First off, i have read and learnt so much from all your posts both in verbal and quant, i really appreciate your time and effort for this too.

Also, i didn't mention this in my previous post but i didn't ask Bunuel for alternate solution just because i found other solutions as 'complicated or long or i wanted a short-cut method or solutions looks hard', i actually solved this question and arrived at the correct answer in the same way as you and others did.

The reason i asked for a alternate solution is because as we all know Bunuel and other experts too often come up with a solution which is completely different and less time consuming. That was exactly my reason behind asking for an alternate solution. Also the post was around 3 years old so was hoping if someone may have some new information on this.

Probability is one of the challenging topics for me, even though i was able to solve this problem i felt i took more time to solve it than that is necessary and i just wanted to improve on it by taking some help.

Thanks again for your advise and insights on facing problems like these.

Re: A fair coin is tossed 6 times. What is the probability of getting no [#permalink]

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21 Dec 2016, 06:54

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A fair coin is tossed 6 times. What is the probability of getting no [#permalink]

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05 Jan 2017, 11:38

Here is the method that I used.

1) Number of outcomes with 6T = 1 2) Number of outcomes with 5T +1H = 6!/5! = 6 3) Number of outcomes with 4T +2H = 6!/(4!*2!) = 15 [But this includes combinations where the 2Hs are together] i.e. [HH] is placed in any of the spaces (_) in _T_T_T_T_ So, subtracting 5 from (3). 4) Number of outcomes with 3T +3H = 6!/(3!*3!) = 20 [Here again, we have to subtract combinations where the 2Hs are together] i.e. [HH] is placed in any of the spaces (_) in {[_T_T_T_H], [_T_T_HT_], [_T_HT_T_], [_HT_T_T_]}. Realize I've removed one space (_) after the each H. So, subtracting 4*4 = 16. Above will take care of all combinations where 3Hs are together.

Total favourable combinations = 1+6+15-5+20-16 = 21 P = 21/64 A.

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