Afair coin is tossed n times. what is the probability that no two consecutive heads appear? a fair coin is tossed n times. what is the probability that no two consecutive heads appear?

If a coin is tossed, for example, 10 times:

There are 2^10 = 1024 total possible outcomes of 10 tosses. 

Now we count the number of possible outcomes with no two consecutive heads. Let x_n be this number for n tosses - we wish to find x_10. For n > 2, the string of tosses must fall into one of two disjoint categories: 

(i) tosses ending with a tail 

(ii) tosses ending with a tail then a head 

(i.e. tosses ending with a head then a head is impossible) 

Given that the first n-1 or n-2 tosses must also have no two consecutive heads, we sum the number of possibilities of case (i) and (ii) and find 
x_n = x_{n-1} + x_{n-2}. 

We also easily see that x_1 = 2, x_2 = 3. From the above recurrence we can compute x_n as terms of the Fibonacci sequence until x_10 = 144. 

Since the coin is unbiased all these possibilities are equally likely and the answer is 144/1024 = 9/64.

The answer to #44 is 10.4444, so g is the answer i divided the time by the miles. 47

Two because two of one thing is two.