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(#M40121319) COCUBES question probability Keep an EYE

probability that 2 heads do not occur consecutively when a fair coin is tossed 10 times?

Asked In cocubes Shintala Divya (10 years ago)

Unsolved Read Solution (15)

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Read Solution (15) :

Mau Chakravorty  (9 years ago)

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

Is this solution Helpfull? Yes (27) | No (7)  

Bhuwan Pandey  (9 years ago)

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=0.140625

Is this solution Helpfull? Yes (7) | No (1)  

Prashant   (10 years ago)

144/1024
n+2 fibonacci number/2^n

Is this solution Helpfull? Yes (6) | No (5)  

leela  (10 years ago)

5/10 = 1/2

Is this solution Helpfull? Yes (4) | No (22)  

anand  (9 years ago)

144/1024
fibonacci sum till 10 times by total posibilty which is 2^10

Is this solution Helpfull? Yes (4) | No (3)  

pundlik  (9 years ago)

for first toss gives head
then P(H)=1/2

and second toss must be of probability of 1 for tail.
after that it must be 1 for head and goes on.

so for any number of tosses probability for getting consecutive head = 1/2

Is this solution Helpfull? Yes (2) | No (13)  

Amit Kumar Singh  (9 years ago)

from the question
its shows that for every time the possible probability is (1/2).
lets create a sample space {H/T, T, H/T, T, H/T, T, H/T, T, H/T, T} [reason: as question says that no 2 heads, H continuous]
now its easy,
P(getting Heads) = 5*(1/2) = 5/2
P(getting Tails) = 5*(1/2)+ 5 = 15/2
so, P(total no 2 Heads together) = (5/2)/(15/2) = 1/3 = 0.33

Is this solution Helpfull? Yes (2) | No (5)  

milan gupta  (9 years ago)

144/2^10=9/64

Is this solution Helpfull? Yes (1) | No (3)  

harini  (9 years ago)

can anybody give more explanation

Is this solution Helpfull? Yes (1) | No (1)  

r.harini  (9 years ago)

pls somebody give correct explanation for this sum

Is this solution Helpfull? Yes (1) | No (4)  

aditi roy  (8 years ago)

P(COMPLEMENT OF HH)=1-P(HH)=1-1/(2^10)=1023/1024

Is this solution Helpfull? Yes (1) | No (1)  

Deepak prajapati  (10 years ago)

1/2*1/2=1/4

Is this solution Helpfull? Yes | No (7)  

Salman khan  (8 years ago)

the sample space is 2^10.
since two heads should not occur consecutively so we should first check when these occur consecutively
there are possible 9 cases when heads occur
after that we should check all possible combination after two heads that are 10C8
since the 2 heads do not occur consecutively , this would be given by (All - 9) ie (10C8 -9)
p(E) = (10C8 - 9)/2^10

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Janani  (8 years ago)

guys any one correct answer pls....

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prabhu  (8 years ago)

someone plzz send me capgemini placement papers jillaprabhu123@gmail.com

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