A fair coin is tossed 10 times. What is the probability that

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:

3/4; 5/8; 8/16; 13/32; 21/64; 34/128; 55/256; 89/512; 144/1028...

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

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)

For 0 Heads 1 way

Total Fav events = 2+10+20+20+10+1

Probability=63/2^10

ANS --> D

1/2^4-1/2^10 is not correct

I should also take into account variants like x00x00x0x

Therefore p>1/2^4

There's nothing saying that half the tosses have to be tails. You could have

HHHHHHHHHH
HTHHHHHHHH
HTHTHHHHHH
HHHHHHHTHH
HTHTHTHHHT
etc etc

there are tons and tons of combinations possible.

I don't think that is 2-minutes GMAT problem to solve it. I guess we should find quick way to say D, but do not calculate probability.

No 2 heads should be together meaning Heads should be between 2 tails. Hence there are 2 ways Heads can be obtained:

1. Places 1,3,5,7,9
2. Places 2,4,6,8,10

From above 5/10 = 1/2

Then the probability of each of the 5 positions is 1/2*1/2*1/2*1/2*1/2 = 1/32

therefore, 1/2 * 1/32 = 1/64

Either 1 or 2 will occur

Hence 1/64 + 1/64 = 2/64 = 1/32

Ans 1/2^5

C

What is OA?

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!/(10-1)!*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 equation

The 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)^(10-144) =
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

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.

My answer is C .
i.e. 1/2^5.

Not occuring consecutively is as good as saying that 'Head' appear only 5 times out of the 10 toss.


What is the OA?

D But I'm not sure.

1. we have two patterns:

x-head
0-tail

x0x0x0x0x0
0x0x0x0x0x

for each pattern the probability is p0=1/2^10

2. try to change one by one "x" on "0".

p=2*p0+2*p0*5C1+2*p0*5C2*+2*p0*5C3+2*p0*5C4+p0*5C3=
p0*(2+10+20+20+10+1)=63*p0=2^6-1/2^10=1/2^4-1/2^10

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?

I've used Monte-Carlo method on my PC and obtained p~0.1406

OA given was "B". But that is not correct at all. And so it seems, by looking at your answers. This was really really tough!

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.