Problem Solving

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)1/2^6

E)None of the above

cindybrown wrote:9 . 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)1/2^6

E)None of the above

I saw this question some time back in some other forum. Its a bit tricky.
Lets try to understand it with basic approach.


We have to consider 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.
...

Looking above results , we can see the pattern in "no consecutive H": 2, 3, 5, 8...

It looks like a Fibonacci series 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=.140625

IMO "E"

cindybrown wrote:9 . 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)1/2^6

E)None of the above

hi its a good question and to solve it we have to consider the following case..!!
Case 1: all tails
1 case
CASE 2:
9T 1H
total such cases 10
CASE 3:
8T 2H
XTXTXTXTXTXTXTXTX

Here X represents the places where Head can occur..!!
we can choose any 2 places frm 9 X = 9C2 =36
CASE 4:
7T 3H
XTXTXTXTXTXTXTX
we can choose any 3 places for H frm 8X = 8C3=56
CASE 5:
6T 4H
by same logic we get 7C4 cases = 35
CASE 6:
5H 5T
by same logic we get 6C5 = 6
so total cases 1 + 10 +36 +56 +35 +6= 144 cases
So answer shd be 144/2^10

hence E

i hope it helps..!!!

i agree with 6983manish and manpsingh87. not repeating the solution...got the same answer. E

@thanks everyone. Is there a easier way to do this question? how can i do this in 2 min????

Let's look at this as a whole question - with answer choices.

if you toss a coin ten time then then nubmer of possible sets of tosses is 2^10? now for probability you have to figure out how many of those will be times when you have two consecutive heads - which the others have done very well. But there is a time limit and if you aren't fast enough then well, what can you do. Let's look at our answers:

if you know 2^10 must be the denominator then think about what the numerator could be - if the answer is 1/2^4 then the numerator must have been 2^6 = which is 4^3 which is 64 - that is really very low for all the ways you can't have two heads - if you play with the pattern a little you see that you can get that high pretty quickly - for instance - 1 way with all tails, 10 ways with 1 head,8 ways with head first, tail second, similarly 8 ways for tail last and we are already up to 27 and we haven't conquered the middle yet so you may be safe choosing E just based on the idea of pattern and not have to solve.

Of course, if you can solve it would be best but not everyone can solve every problem.

please let me know the 10th element of the fibonacci series

It looks like a Fibonacci series of sequence and it will continue: 2, 3, 5, 8....

as its not posisble to manually find the 10th elements in exam....

manpsingh87 wrote:

cindybrown wrote:9 . 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)1/2^6

E)None of the above

hi its a good question and to solve it we have to consider the following case..!!
Case 1: all tails
1 case
CASE 2:
9T 1H
total such cases 10
CASE 3:
8T 2H
XTXTXTXTXTXTXTXTX

Here X represents the places where Head can occur..!!
we can choose any 2 places frm 9 X = 9C2 =36
CASE 4:
7T 3H
XTXTXTXTXTXTXTX
we can choose any 3 places for H frm 8X = 8C3=56
CASE 5:
6T 4H
by same logic we get 7C4 cases = 35
CASE 6:
5H 5T
by same logic we get 6C5 = 6
so total cases 1 + 10 +36 +56 +35 +6= 144 cases
So answer shd be 144/2^10

hence E

i hope it helps..!!!

Why did u stop at 6c5....can u plz explain

ruplun wrote:

manpsingh87 wrote:

cindybrown wrote:9 . 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)1/2^6

E)None of the above

hi its a good question and to solve it we have to consider the following case..!!
Case 1: all tails
1 case
CASE 2:
9T 1H
total such cases 10
CASE 3:
8T 2H
XTXTXTXTXTXTXTXTX

Here X represents the places where Head can occur..!!
we can choose any 2 places frm 9 X = 9C2 =36
CASE 4:
7T 3H
XTXTXTXTXTXTXTX
we can choose any 3 places for H frm 8X = 8C3=56
CASE 5:
6T 4H
by same logic we get 7C4 cases = 35
CASE 6:
5H 5T
by same logic we get 6C5 = 6
so total cases 1 + 10 +36 +56 +35 +6= 144 cases
So answer shd be 144/2^10

hence E

i hope it helps..!!!

Why did u stop at 6c5....can u plz explain

he had taken 5h 5t.
that is th limit
bcoz for 4tails will have only 5 spaces to fill but the heads are 6. hope u got it