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Re: ProbabilityCoin Toss
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22 Oct 2009, 12:42
The way I solved this was a bit tedious, but it worked.
1. Construct a tree diagram with heads and tails out to 5 decisions 2. Find the number of decisions that will have no consecutive heads for each leading branch of the decision tree to develop a pattern. You will end up getting 2 on the first toss, 3, 5, 8, 13, etc. 3. Recognize that this is the fibonacci sequence, and simply go out to n=10, so you get 144 4. 144/total possibilities = 144/2^10 = 9/2^6, Answer D
The problem with this method is that it takes time to develop the pattern. 2,3,5,8 might not be fibonacci because the next number could be 12. 2,3,5,8,13 has to be fibonacci, though. So you need to go out pretty far on the tree to get a good answer. Probably not possible in under 2 minutes unless you use the method mentioned above by another poster.



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Re: ProbabilityCoin Toss
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Updated on: 14 Nov 2009, 20:49
I was able to complete this problem by thinking about the binomial distribution concept. Prework: One of the first things taught in any probability math class on the this subject and when coin flipping is introduced is the fact that if you have 5 flips, and of those 5 you want 1 to be a heads, then you have _T_T_T_T_ different spots in which to place the head. It can come before the first tail, after the last tail, or in the middle. This concept MUST be understood on the GMAT. It is one of the most basic/first things we learn in probability math. Read my 2nd post for how I ended up doing the work.
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Originally posted by benjiboo on 12 Nov 2009, 13:43.
Last edited by benjiboo on 14 Nov 2009, 20:49, edited 4 times in total.



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Re: ProbabilityCoin Toss
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13 Nov 2009, 11:56
benjiboo wrote: This is the binomial distribution concept.
2: We must reword as with most GMAT problems. We are really being asked about the probability of 0,1,2,3,4, or 5 heads (successes) all within the one 10trail run.
This is not a straightforward binomial distribution question. You cannot rephrase the question as you've done; we do not want to count all of the ways we might get five Heads, because sometimes when we get 5 Heads, two or more Heads will occur consecutively. We do not, for example, want to count the sequence HHHTTTHHTT. If you complete the calculation as you suggest, you'd find the answer to be 319/512, or 62.3%, which is considerably too high. And I'm not sure where the jpeg you posted comes from, but it contains a mistake; when the author refers to \(p\), he or she means to refer to \(\pi\).
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Re: ProbabilityCoin Toss
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13 Nov 2009, 13:02
IanStewart wrote: benjiboo wrote: This is the binomial distribution concept.
2: We must reword as with most GMAT problems. We are really being asked about the probability of 0,1,2,3,4, or 5 heads (successes) all within the one 10trail run.
This is not a straightforward binomial distribution question. You cannot rephrase the question as you've done; we do not want to count all of the ways we might get five Heads, because sometimes when we get 5 Heads, two or more Heads will occur consecutively. We do not, for example, want to count the sequence HHHTTTHHTT. If you complete the calculation as you suggest, you'd find the answer to be 319/512, or 62.3%, which is considerably too high. And I'm not sure where the jpeg you posted comes from, but it contains a mistake; when the author refers to \(p\), he or she means to refer to \(\pi\). I agree this is not a straight forward binomial distribution question. I think the formula needs to be modified for a question like this, which I tried to do. I am sure there is a mathematical answer to this using some form of a maniuplated binomial distribution equation. Lets discuss this together. The question is asking you the number of places you can place a success given (0 heads given 10 tails) + (1 heads given 9 tails) + (2 heads 8 tails) + (3 heads 7 tails) + (4 heads 6 tails) + (5 heads 5 tails) in 10 trials of a binomial event, whereas a success is defined as a heads in any spot that is not next to another heads... Meaning, for (1 head given 9 tails) you have _T_T_T_T_T_T_T_T_T_ 10 spots. You must take the 10 spots and choose 1 of them. How many different ways can you choose that 1 spot? There are 10c1 ways. Or 10!/(101)!*1!.] As you see in the part you quoted of me, I said the "probability of getting 0,1,2,3,4, or 5 heads (successes)" Therefore, the wording is correct, as I am eliminating the failures by not counting spots for them. Also, if you do correctly use the equation I provided (my explanation tells you that in a case like this the first part of the equation is the SUM of the first part for each success), you get the right answer. Here is the math: First half of the equation:0 heads, 10 tails= TTTTTTTTTT = 0c0 1 Heads, 9 tails = _T_T_T_T_T_T_T_T_T_ = 10c1 2 heads, 8 tails = _T_T_T_T_T_T_T_T_ = 9c2 3 heads, 7 tails = _T_T_T_T_T_T_T_ = 8c3 4 heads, 6 tails = _T_T_T_T_T_T_ = 7c4 5 heads, 5 tails = _T_T_T_T_T_ = 6c5 0c0 + 10c1 + 9c2 + 8c3 + 7c4 + 6c5 = 1 + 10 + 36 + 56 + 35 + 6 = 144 second part of the equationThe second part of the formula, r becomes 144, and N is 10. Why is r 144? Because as from above there are 144 chances of a success.(1/2)144 * (1/2)^(10144) = 1/2^144 * 1/2^134 = 1/2^144 * 2^134 = 2^10 = 1/2^10 = 1/1024 Put it together
144 * 1/1024 = 144/1024 = answer 144/1024 = probability of getting 0,1,2,3,4, or 5 heads in a successful manor as defined by the question.  I believe this is right. However I welcome all discussion. My goal is to provide good and clear explanations for forum members. Sometimes somebody has already given a great explanation, so I try and find an alternative explanation to help further people's understandings.  Benjiboo



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Re: ProbabilityCoin Toss
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13 Nov 2009, 13:43
benjiboo wrote: 0 heads, 10 tails= TTTTTTTTTT = 0c0 1 Heads, 9 tails = _T_T_T_T_T_T_T_T_T_ = 10c1 2 heads, 8 tails = _T_T_T_T_T_T_T_T_ = 9c2 3 heads, 7 tails = _T_T_T_T_T_T_T_ = 8c3 4 heads, 6 tails = _T_T_T_T_T_T_ = 7c4 5 heads, 5 tails = _T_T_T_T_T_ = 6c5
0c0 + 10c1 + 9c2 + 8c3 + 7c4 + 6c5 = 1 + 10 + 36 + 56 + 35 + 6 = 144
Yes, that's the solution KASSALMD gave in an earlier post (though, as I later pointed out, he omitted the cases of 0 Heads and 1 Head), and it is correct. There's really no need to use a binomial probability formula here  by the definition of probability, we have 1 + 10C1 + 9C2 + 8C3 + 7C4 + 6C5 ways of getting the result we want, and 2^10 possible outcomes in total, so dividing the first number by the second gives our answer. There would only be an advantage to using binomial probability if the probability of getting Heads was different from 1/2.
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Re: ProbabilityCoin Toss
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03 Sep 2010, 09:38
parsifal wrote: LM wrote: A fair coin is tossed 10 times. What is the probability that two heads do not occur consecutively?
A) 1/ 2^4
B) 1/2^3
C)1/2^5
D) None of the above Let's arrange the possibilities: 1. Tosses 1,3,5,7,9 are tails The others can be either heads or tails: 2^5 possibilities. 2. Tosses 2,4,6,8,10 are tails The others can be either heads or tails: 2^5 possibilities. Hence, total possibilites where no 2 consecutive are heads is: 2 * 2^5 Probability = 2^6/2^10 = 1/2^4 what if in case 1, all the tosses are tails as well as in case 2. therefore a possibility of all the tosses are tails is counted twice. my point is each possibility must only be counted once not twice. I am not sure though. pls correct me if i am wrong.



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Re: ProbabilityCoin Toss
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16 Mar 2011, 10:35
Yeah I got the same answer, D specifically 144/1024 But I did it as 1probability(getting 2 Heads in a row) I think it was more complicated...but still used fibonacci
2^8+(2^7)*1+(2^6)*2+(2^5)*3+(2^4)*5+(2^3)*8+(2^2)*13+(2^1)*21+(2^0)*34 = 880
Notice the 1, 2, 3, 5, 8, 13, 21, 34 sequence
1880/1024 = 144/1024 WOw, spent the whole night on this one.



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Re: ProbabilityCoin Toss
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23 Feb 2012, 06:54
Bunuel wrote: A fair coin is tossed 10 times. What is the probability that two heads do not occur consecutively? A) 1/ 2^4 B) 1/2^3 C) 1/2^5 D) None of the above
OK, here is my solution:
Possible number of patterns (total number of combinations) 2^n (each time either H or T=2 outcomes, 10 times=2^n).
Let's check two consecutive H: If we toss once we'll have 2^1=2 combinations: H, T  2 outcomes with NO 2 consecutive H. If we toss twice we'll have 2^2=4 combinations: HT, TH, TT, HH  3 outcomes with NO 2 consecutive H. If we toss 3 times we'll have 2^3=8 combinations: TTT, TTH, THT, HTT, HTH, HHT, THH, HHH 5 outcomes with NO 2 consecutive H. If we toss 4 times we'll have 2^4=16 combinations:... 8 outcomes with NO 2 consecutive H. ...
On this stage we can see the pattern in "no consecutive H": 2, 3, 5, 8...
I guess it's Fibonacci type of sequence and it will continue: 2, 3, 5, 8, 13, 21, 34, 55, 89, 144.
144 is outcomes with no consecutive H if we toss 10 times.
P(no two consecutive H in 10 toss)=144/2^10=144/1024=14.0625% This is great as usual from you
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Re: A fair coin is tossed 10 times. What is the probability that
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