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I never said that half the tosses be tails.
However, I did say that ATLEAST HALT must be tails. The other half may be heads or tails.
This will ensure that 2 heads dont occur consecutively.
In the examples that you have given, 2 heads occur consecutively. Note that 2 OR MORE consecutive also implies 2 consecutive.

I never said that half the tosses be tails. However, I did say that ATLEAST HALT must be tails. The other half may be heads or tails. This will ensure that 2 heads dont occur consecutively. In the examples that you have given, 2 heads occur consecutively. Note that 2 OR MORE consecutive also implies 2 consecutive.

my mistake, I mixed up tails and heads and didn't re-read the question.

I understand your approach now. Is there an OA for this question?

Re: Probability-Coin Toss [#permalink]
20 Aug 2008, 17:41

3

This post received KUDOS

Expert's post

KASSALMD wrote:

This is not a hard question. Bear in mind that there could be 2 or 3 or 4 or 5 Heads but no more.

You've left out the possibility that there is one head, or zero heads. If you add those in, you should find there are 144 possibilities, and you'll get the correct answer: 144/2^10 = 9/64.

You can also do the problem inductively, which demonstrates an interesting connection between this problem and Fibonacci numbers. If you flip two coins, there are 3 sequences which do not have two consecutive heads:

TH HT TT

No matter how we get to 10 flips with no consecutive heads, we have to start with one of these three 'words'. If we flip another coin (i.e., if we add an H or a T to the end of one of the three words above), and we must not have two consecutive heads, we can:

-only add a T to the end of a word that ends in H; -add either a T or an H to the end of a word that ends in T.

So, if we have:

x + y words in total with n letters, consisting of x words that end in H y words that end in T

we can make x + 2y words in total with (n+1) letters (because from each word ending in T we can make two new words, and from each ending in H we can only make one), and of these, we'll have: y words that end in H x + y words that end in T

and then can make 2x + 3y words in total with (n+2) letters x + y words that end in H x + 2y words that end in T

and so on.

Notice, though, if you let \(a_1 = x \\ a_2 = y \\ a_n = a_{n-2} + a_{n-1} \\ \text{then} \\ a_3 = x+y \\ a_4 = x + 2y \\ a_5 = 2x + 3y\)

and so on. That is, the number of words you can make are just numbers from the Fibonacci sequence, since a_1 = 1 and a_2 = 2. If you did this problem for any number of coins, the numerators would all be Fibonacci numbers. So for 2, 3, 4, 5, 6, 7, 8, 9, and 10 coins, the answers will be:

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Re: Probability-Coin Toss [#permalink]
21 Aug 2008, 04:00

I think the key to this question is realising (on the 120th second!) that the denominator has to be 2^10. Then you pick answer D and move on... I must confess I did not do it that way though. _________________

Re: Probability-Coin Toss [#permalink]
21 Aug 2008, 06:07

Expert's post

Nerdboy wrote:

I think the key to this question is realising (on the 120th second!) that the denominator has to be 2^10. Then you pick answer D and move on.

While that is sometimes a very useful technique, it doesn't help much here. Yes, it's true that there is a total of 2^(10) outcomes, but if the numerator is even, there will be cancellation. All we can be sure of is that the denominator of the correct answer is a factor of 2^(10). Indeed, after canceling, you find that the denominator of the correct answer is 2^6. _________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Re: Probability-Coin Toss [#permalink]
21 Aug 2008, 10:27

2

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here is my analysis :

Lets start with the maximum count of heads possiblle i.e 5

For 5 Heads there can 2 possible ways ; THTH...... OR HTHT.....

For 4 Heads there can be 5*2 ways ; for this first lets arrange 5 Tails in all alternate positions , now we are left with one Tail which can come in 5 empty spaces so 5 options for that , but again we can start like THTH...... OR HTHT... therefore 5*2

For 3 Heads ; first lets arrange 5 Tails in all alternate positions (as this gives us the condition that no 2 Heads can come together ) now for the othet 2 Tails we have 5 spaces so 5C2 possible ways , again THTH... OR HTHT... so 5C3*2

For 2 Heads ; 5C3*2

For 1 Head 10 different ways (another way of looking at it : 5C4*2)

Re: Probability-Coin Toss [#permalink]
20 Oct 2009, 11:24

10

This post received KUDOS

Expert's post

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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, HHH5 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% _________________

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