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

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

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

I should also take into account variants like x00x00x0x

Therefore p>1/2^4

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

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 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 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.

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...

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.

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.

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

D! but I'm not sure

My method:

main premise- there could'nt be more than 5 H.

1) zero H--> 10c0 * (1/2)^10= (1/2)^10

2) one H --> 10c1 * (1/2)^10 = 10*(1/2)^10

3) two H-->10c2 * (1/2)^10 * (prob that adjacent)=10c2 * (1/2)^10*(1/5)= 9*(1/2)^10

4) 3 H--> 10c3*(1/2)^10*(1/10)=12*(1/2)^10

5) 4 H--> 10c4*(1/2)^10*(1/30)=7*(1/2)^10

6) 5 H--> 10c5**1/2)^10*(1/42)=6*(1/2)^10

the sum of the probabilities is 45*(1/2)^10 = 0.4395
thus, my answer is D

Very hard!
I used Excel to list all probabilities. The answer is 144/1028 = 9/64 = 0.1406

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?