Given a fair coin which is tossed **N** times, the task is to determine the probability such that no two heads occur consecutively.

**Examples:**

Input:N = 2

Output:0.75

Explanation:

When the coin is tossed 2 times, the possible outcomes are {TH, HT, TT, HH}.

Since in 3 out of 4 outcomes, heads don’t occur together.

Therefore, the required probability is (3/4) or 0.75

Input:N = 3

Output:0.62

Explanation:

When the coin is tossed 3 times, the possible outcomes are {TTT, HTT, THT, TTH, HHT, HTH, THH, HHH}.

Since in 5 out of 8 outcomes, heads don’t occur together.

Therefore, the required probability is (5/8) or 0.62

**Approach:** The following observation on the number of favourable outcomes has to be made.

**When N = 1:**The possible outcomes are {T, H}. There are**two**favourable outcomes out of two.**When N = 2:**The possible outcomes are {TH, HT, TT, HH}. There are**three**favourable outcomes out of four.**When N = 3:**Similarly, the possible outcomes are {TTT, HTT, THT, TTH, HHT, HTH, THH, HHH}. There are**five**favourable outcomes out of eight.**When N = 4:**Similarly, the possible outcomes are {TTTT, TTTH, TTHT, THTT, HTTT, TTHH, THTH, HTHT, HHTT, THHT, HTTH, THHH, HTHH, HHTH, HHHT, HHHH}. There are**eight**favourable outcomes out of sixteen.

Clearly, the number of favourable outcomes follow a Fibonacci series where Fn(1) = 2, Fn(2) = 3 and so on. Therefore, the idea is to implement the Fibonacci sequence in order to find the number of favourable cases. Clearly, the total number of cases are 2^{N}.

To calculate the probability, the following formula is used:

P = favourable cases / Total number of cases

Below is the implementation of the above approach:

`# Python implementation to find the ` `# probability of not getting two ` `# consecutive heads together when ` `# N coins are tossed ` ` ` ` ` `import` `math ` ` ` `# Function to compute the N-th ` `# Fibonacci number in the ` `# sequence where a = 2 ` `# and b = 3 ` `def` `probability(N): ` ` ` ` ` `# The first two numbers in ` ` ` `# the sequence are initialized ` ` ` `a ` `=` `2` ` ` `b ` `=` `3` ` ` ` ` `# Base cases ` ` ` `if` `N ` `=` `=` `1` `: ` ` ` `return` `a ` ` ` `elif` `N ` `=` `=` `2` `: ` ` ` `return` `b ` ` ` `else` `: ` ` ` ` ` `# Loop to compute the fibonacci ` ` ` `# sequence based on the first ` ` ` `# two initialized numbers ` ` ` `for` `i ` `in` `range` `(` `3` `, N ` `+` `1` `): ` ` ` `c ` `=` `a ` `+` `b ` ` ` `a ` `=` `b ` ` ` `b ` `=` `c ` ` ` `return` `b ` ` ` `# Function to find the probability ` `# of not getting two consecutive ` `# heads when N coins are tossed ` `def` `operations(N): ` ` ` ` ` `# Computing the number of ` ` ` `# favourable cases ` ` ` `x ` `=` `probability (N) ` ` ` ` ` `# Computing the number of ` ` ` `# all possible outcomes for ` ` ` `# N tosses ` ` ` `y ` `=` `math.` `pow` `(` `2` `, N) ` ` ` ` ` `return` `round` `(x ` `/` `y, ` `2` `) ` ` ` `# Driver code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` ` ` `N ` `=` `10` ` ` ` ` `print` `(operations(N)) ` ` ` |

*chevron_right*

*filter_none*

**Output:**

0.14

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