A fair coin is tossed

[cindybrown]

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

[6983manish]

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"

[manpsingh87]

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..!!!

[force5]

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

[cindybrown]

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

[tpr-becky]

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.

[ruplun]

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....

[ruplun]

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

[naveen451]

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