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22 Oct 2009, 12:42
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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.
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.
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.
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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bionomail distro.jpg [ 82.67 KiB | Viewed 50697 times ]
13 Nov 2009, 11:56
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benjiboo
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 \(\pi\).
