A coin is tossed M+N times,M>N,then the chance of getting at least consecutive head is
To find the chance of getting at least one consecutive head when a coin is tossed M+N times, where M > N, we can use the principle of inclusion-exclusion.
The total number of possible outcomes when a coin is tossed M+N times is 2^(M+N), as each toss has 2 possible outcomes (head or tail) and the tosses are independent.
Now, let's calculate the number of sequences where there are no consecutive heads. We can break this down into two cases:
Case 1: The last toss is heads
In this case, we have M-1 tosses remaining, and we cannot have any consecutive heads in these M-1 tosses. So, the number of such sequences is equal to the number of sequences where there are no consecutive heads in M-1 tosses, which is 2^(M-1).
Case 2: The last toss is tails
In this case, we have M tosses remaining, and we cannot have any consecutive heads in these M tosses. So, the number of such sequences is equal to the number of sequences where there are no consecutive heads in M tosses, which is 2^M.
Now, by the principle of inclusion-exclusion, the number of sequences without any consecutive heads is given by:
2^(M-1) - 2^M
Finally, the chance of getting at least one consecutive head is given by the complement of getting no consecutive heads, so:
P(at least one consecutive head) = 1 - (2^(M-1) - 2^M) / 2^(M+N)
Simplifying the expression further will give you the final probability.