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

• 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
• 9 years agoHelpfull: Yes(27) No(7)
• 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
• 8 years agoHelpfull: Yes(7) No(1)
• 144/1024
  n+2 fibonacci number/2^n
• 9 years agoHelpfull: Yes(6) No(5)
• 5/10 = 1/2
  
• 9 years agoHelpfull: Yes(4) No(22)
• 144/1024
  fibonacci sum till 10 times by total posibilty which is 2^10
• 9 years agoHelpfull: Yes(4) No(3)
• 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
• 9 years agoHelpfull: Yes(2) No(13)
• 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
• 8 years agoHelpfull: Yes(2) No(5)
• 144/2^10=9/64
  
• 9 years agoHelpfull: Yes(1) No(3)
• can anybody give more explanation
• 9 years agoHelpfull: Yes(1) No(1)
• pls somebody give correct explanation for this sum
  
• 9 years agoHelpfull: Yes(1) No(4)
• P(COMPLEMENT OF HH)=1-P(HH)=1-1/(2^10)=1023/1024
• 7 years agoHelpfull: Yes(1) No(1)
• 1/2*1/2=1/4
• 9 years agoHelpfull: Yes(0) No(7)
• 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
• 8 years agoHelpfull: Yes(0) No(0)
• guys any one correct answer pls....
• 8 years agoHelpfull: Yes(0) No(0)
• someone plzz send me capgemini placement papers jillaprabhu123@gmail.com
• 8 years agoHelpfull: Yes(0) No(0)