online examination

Question

if 8 coins are tossed then find the probability that no two heads appear consecutively??

Comments

This is one of the standard problems and has a trick to solve :

F_n+2 / 2ⁿ    (Where F_n is n^th Fibonacci number)

As n=8 here

F₁₀ / 2⁸ = 55 / 256

Answer 1

if 8 coins are tossed then

1) 1 Head , 7 Tails

x T x T x T x T x T x T x T x 

Head can be sit in any of 8 position marked by "x"  i.e = 8P1/1! = 8 ways

 2) 2 Head , 6 Tails

x T x T x T x T x T x T x 

similarly , 7P2/2! =21 ways

  3) 3 Head , 5 Tails

x T x T x T x T x T x 

similarly , 6P3/3! = 20 ways

  4) 4 Head , 4 Tails

x T x T x T x T x 

similarly , 5P4/4! = 5 ways

now, 5 Head and 3 Tails is not possible because consecutive head will occur.

one fav. case is all 8 Tails = 1 way

total = 55 ways 

prob. = 55/2^8 =55/256

Comments

0 head and 8 Tails is also favourable?

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in 0 head and 8 Tails there is no two consecutive heads