Let $f_n$denote the number of ways of tossing a coin n times such that successive heads never appear. Argue that$f_n=f_{n-1}+f_{n-2},n\ge 2$, where$f_0=1,f_1=2$.

Hint: How many outcomes are there that start with ahead, and how many start with a tail? If $P_n$denotes the probability that successive heads never appear when a coin is tossed n times, find $p_n$(in terms of $f_n$) when all possible outcomes of the $n$tosses are assumed equally likely. Compute$P_{10}.$

Let $f_n$denote the number of ways of tossing a coin $n$times such that successive heads never appear.

$n$coin tosses, note which side the coin lands on.

$f_n$- the number of ways in which the experiment can end so that there are no two consecutive heads.

Count $f_n$using the hint: The total number of results of the experiment in which there are no consecutive heads is the sum of the number of results with no two consecutive heads if the first toss ended ahead and the number of such results if the first toss resulted in tails.

If the first toss was head, the next should be tails so that there are no two consecutive heads. And the remaining $n-2$tosses can be any such that no two heads are consecutive, and the number of such results is$f_{n-1}$.

If the first toss was tails, the remaining $n-1$tosses can be any such that there are no two consecutive heads, and by the definition of$f$, the number of such results is$f_{n-1}$.

So$f_n=f_{n-1}+f_{n-2}$

Naturally, $f_0=1f_1=2$

Because $2^n$is the number of results of a series of $n$two outcome experiments.

By iterating the recursion we get$f_{10}=144$, thus

$P_{10}=\frac{144}{1024}\approx 0.14$